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56. Merge Intervals
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‎markdown/56. Merge Intervals.md

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### [56\. Merge Intervals](https://leetcode.com/problems/merge-intervals/)
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Difficulty: **Medium**
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Given a collection of intervals, merge all overlapping intervals.
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**Example 1:**
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```
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Input: [[1,3],[2,6],[8,10],[15,18]]
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Output: [[1,6],[8,10],[15,18]]
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Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
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```
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**Example 2:**
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```
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Input: [[1,4],[4,5]]
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Output: [[1,5]]
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Explanation: Intervals [1,4] and [4,5] are considered overlapping.
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```
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**NOTE:** input types have been changed on April 15, 2019\. Please reset to default code definition to get new method signature.
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#### Solution B
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Language: **Java**
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```java
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class Solution {
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public int[][] merge(int[][] intervals) {
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List<int[]> result = new ArrayList<>();
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if (intervals == null || intervals.length == 0) {
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return result.toArray(new int[result.size()][]);
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}
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Arrays.sort(intervals, Comparator.comparingInt(i -> i[0]));
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// key 是数组的值，value 表示这个是左还是右，左为-1，右为1，如果都有，则求和
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Map<Integer, Integer> map = new TreeMap<>();
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for (int i = 0; i < intervals.length; i++) {
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int left = intervals[i][0];
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if (map.containsKey(left)) {
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map.put(left, map.get(left) - 1);
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} else {
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map.put(left, -1);
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}
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int right = intervals[i][1];
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if (map.containsKey(right)) {
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map.put(right, map.get(right) + 1);
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} else {
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map.put(right, 1);
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}
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}
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int[] interval = new int[2];
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interval[0] = intervals[0][0];
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int i = 0;
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for (Integer integer : map.keySet()) {
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if (i == 0) {
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interval = new int[2];
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interval[0] = integer;
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}
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i += map.get(integer);
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if (i == 0) {
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interval[1] = integer;
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result.add(interval);
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}
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}
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return result.toArray(new int[result.size()][]);
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}
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}
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```
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#### Solution B
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Language: **Java**
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```java
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class Solution {
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public int[][] merge(int[][] intervals) {
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if (intervals.length <= 1)
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return intervals;
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// Sort by ascending starting point
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Arrays.sort(intervals, (i1, i2) -> Integer.compare(i1[0], i2[0]));
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List<int[]> result = new ArrayList<>();
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int[] newInterval = intervals[0];
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result.add(newInterval);
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for (int[] interval : intervals) {
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if (interval[0] <= newInterval[1]) // Overlapping intervals, move the end if needed
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newInterval[1] = Math.max(newInterval[1], interval[1]);
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else { // Disjoint intervals, add the new interval to the list
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newInterval = interval;
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result.add(newInterval);
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}
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}
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return result.toArray(new int[result.size()][]);
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}
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}
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```
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![](http://ww2.sinaimg.cn/large/006tNc79ly1g4xeakhdjej31010u07bt.jpg)

‎src/main/java/leetcode/_56_/Main.java

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package leetcode._56_;
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/**
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* Created by zhangbo54 on 2019-03-04.
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*/
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public class Main {
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public static void main(String[] args) {
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Solution solution = new Solution();
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int[][] intervals = {{1, 3}, {8, 10}, {15, 18}, {2, 6}};
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// int[][] merge = solution.merge(intervals);
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// System.out.println(merge);
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int[][] intervals2 = {{1, 4}, {1, 4}};
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solution.merge(intervals2);
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}
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}
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‎src/main/java/leetcode/_56_/Solution.java

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package leetcode._56_;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.Comparator;
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import java.util.List;
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import java.util.Map;
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import java.util.TreeMap;
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class Solution {
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public int[][] merge(int[][] intervals) {
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List<int[]> result = new ArrayList<>();
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if (intervals == null || intervals.length == 0) {
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return result.toArray(new int[result.size()][]);
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}
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Arrays.sort(intervals, Comparator.comparingInt(i -> i[0]));
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// key 是数组的值，value 表示这个是左还是右，左为-1，右为1，如果都有，则求和
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Map<Integer, Integer> map = new TreeMap<>();
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for (int i = 0; i < intervals.length; i++) {
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int left = intervals[i][0];
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if (map.containsKey(left)) {
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map.put(left, map.get(left) - 1);
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} else {
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map.put(left, -1);
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}
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int right = intervals[i][1];
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if (map.containsKey(right)) {
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map.put(right, map.get(right) + 1);
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} else {
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map.put(right, 1);
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}
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}
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int[] interval = new int[2];
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interval[0] = intervals[0][0];
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int i = 0;
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for (Integer integer : map.keySet()) {
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if (i == 0) {
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interval = new int[2];
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interval[0] = integer;
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}
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i += map.get(integer);
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if (i == 0) {
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interval[1] = integer;
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result.add(interval);
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}
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}
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return result.toArray(new int[result.size()][]);
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}
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}

‎src/main/java/leetcode/_56_/solution.md

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### [56\. Merge Intervals](https://leetcode.com/problems/merge-intervals/)
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Difficulty: **Medium**
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Given a collection of intervals, merge all overlapping intervals.
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**Example 1:**
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```
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Input: [[1,3],[2,6],[8,10],[15,18]]
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Output: [[1,6],[8,10],[15,18]]
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Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
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```
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**Example 2:**
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```
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Input: [[1,4],[4,5]]
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Output: [[1,5]]
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Explanation: Intervals [1,4] and [4,5] are considered overlapping.
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```
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**NOTE:** input types have been changed on April 15, 2019\. Please reset to default code definition to get new method signature.
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#### Solution B
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Language: **Java**
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```java
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class Solution {
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public int[][] merge(int[][] intervals) {
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List<int[]> result = new ArrayList<>();
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if (intervals == null || intervals.length == 0) {
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return result.toArray(new int[result.size()][]);
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}
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Arrays.sort(intervals, Comparator.comparingInt(i -> i[0]));
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// key 是数组的值，value 表示这个是左还是右，左为-1，右为1，如果都有，则求和
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Map<Integer, Integer> map = new TreeMap<>();
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for (int i = 0; i < intervals.length; i++) {
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int left = intervals[i][0];
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if (map.containsKey(left)) {
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map.put(left, map.get(left) - 1);
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} else {
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map.put(left, -1);
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}
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int right = intervals[i][1];
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if (map.containsKey(right)) {
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map.put(right, map.get(right) + 1);
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} else {
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map.put(right, 1);
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}
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}
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int[] interval = new int[2];
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interval[0] = intervals[0][0];
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int i = 0;
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for (Integer integer : map.keySet()) {
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if (i == 0) {
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interval = new int[2];
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interval[0] = integer;
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}
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i += map.get(integer);
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if (i == 0) {
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interval[1] = integer;
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result.add(interval);
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}
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}
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return result.toArray(new int[result.size()][]);
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}
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}
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```
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#### Solution B
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Language: **Java**
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```java
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class Solution {
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public int[][] merge(int[][] intervals) {
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if (intervals.length <= 1)
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return intervals;
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// Sort by ascending starting point
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Arrays.sort(intervals, (i1, i2) -> Integer.compare(i1[0], i2[0]));
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List<int[]> result = new ArrayList<>();
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int[] newInterval = intervals[0];
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result.add(newInterval);
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for (int[] interval : intervals) {
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if (interval[0] <= newInterval[1]) // Overlapping intervals, move the end if needed
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newInterval[1] = Math.max(newInterval[1], interval[1]);
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else { // Disjoint intervals, add the new interval to the list
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newInterval = interval;
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result.add(newInterval);
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}
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}
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return result.toArray(new int[result.size()][]);
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}
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}
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```
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![](http://ww2.sinaimg.cn/large/006tNc79ly1g4xeakhdjej31010u07bt.jpg)

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