287. Find the Duplicate Number

Author: jiangyong07

src/main/java/joshua/leetcode/array/binarysearch/FindTheDuplicateNumber.java

@@ -0,0 +1,65 @@
+// Copyright 2016 Baidu Inc. All rights reserved.
+
+package joshua.leetcode.array.binarysearch;
+
+/**
+ * 287. Find the Duplicate Number<br>
+ * <p/>
+ * <a href = "https://leetcode.com/problems/find-the-duplicate-number/">leetcode link</a>
+ *
+ * @author Jiang Yong ([email protected])
+ */
+public abstract class FindTheDuplicateNumber {
+
+    /**
+     * Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
+     * <p/>
+     * Note:<br/>
+     * <ul>
+     * <li>You must not modify the array (assume the array is read only).</li>
+     * <li>You must use only constant, O(1) extra space.</li>
+     * <li>Your runtime complexity should be less than O(n2).</li>
+     * <li>There is only one duplicate number in the array, but it could be repeated more than once.</li>
+     * </ul>
+     *
+     * @param nums the input array
+     * @return the duplicated number
+     */
+    public abstract int findDuplicate(int[] nums);
+
+    /**
+     * Binary Search算法。
+     *
+     * 假设不出现重复数字的话，假设数组为 [1,2,3,4], n = 4, 那么在落在任意左开右闭区间(i,j]上的元素的个数即为:j-i个，
+     * 如(2,3]区间上的元素为1个，即为:[3], (2,4]区间上的元素为2个，即为[3],[4]
+     * 那么整体思路如下：
+     * 对区间[1,n+1]不断二分，每次计算落在左半区间和右半区间的元素的个数，如果某个区间的元素个数比区间长度大，说明落在这个区间的存在重复元素。
+     * 这样时间复杂度为o(logn*n)
+     *
+     */
+    public static class Solution1 extends FindTheDuplicateNumber {
+
+        @Override
+        public int findDuplicate(int[] nums) {
+            int low = 1;
+            int high = nums.length - 1;
+            while (low < high) {
+                int mid = (low + high) / 2;
+                int offset = high - mid;
+                for (int num : nums) {
+                    if (num > mid && num <= high) {
+                        offset --;
+                    }
+                }
+                // 落在右半边的元素个数比有半边区间长度大，说明右半边有重复元素，窗口移到右半边
+                if (offset < 0) {
+                    low = mid + 1;
+                } else {
+                    high = mid;
+                }
+            }
+            return low;
+        }
+    }
+
+}

src/main/java/joshua/leetcode/array/greedy/RemoveDuplicateLetters.java

@@ -33,6 +33,9 @@ public abstract class RemoveDuplicateLetters {
     static class Solution1 extends RemoveDuplicateLetters {
 
         /**
+         * 结果要求是按照字母顺序从小到大排列的。
+         * 算法：
+         *
          * 使用栈来保存最后的结果，并用额外空间保存所有的字母的出现次数。
          * 元素进栈时，和栈顶元素比较。
          * 同时需要保存该字母是否已经在结果中的信息。例如
@@ -58,8 +61,10 @@ public String removeDuplicateLetters(String s) {
                     counters.put(ch, counters.get(ch) - 1);
                     continue;
                 }
+                //为ch找一个最靠前的位置，只要栈顶字母比ch大，后面还会再出现元素，就会一直弹出栈顶元素。
                 while (!result.isEmpty()) {
                     char top = result.getLast();
+                    // 如果后面不会再出发top这个字母，则不能出栈了。
                     if (counters.get(top) == 1)
                         break;
                     if (top >= ch) {

src/test/java/joshua/leetcode/array/binarysearch/FindTheDuplicateNumberTest.java

@@ -0,0 +1,31 @@
+package joshua.leetcode.array.binarysearch;
+
+import static org.junit.Assert.assertEquals;
+
+import java.util.Map;
+
+import org.junit.Before;
+import org.junit.Test;
+
+import com.google.common.collect.Maps;
+
+public class FindTheDuplicateNumberTest {
+
+    private Map<int[],Integer> cases = Maps.newHashMap();
+
+    @Before
+    public void setUp() {
+        int[] nums  = new int[]{2, 2, 1, 3,2};
+        cases.put(nums, 2);
+        nums  = new int[]{1, 1, 1, 3,2};
+        cases.put(nums, 1);
+    }
+
+    @Test
+    public void testSolution1() {
+        FindTheDuplicateNumber solution = new FindTheDuplicateNumber.Solution1();
+        for(int[] nums : cases.keySet()) {
+            assertEquals((int)cases.get(nums), solution.findDuplicate(nums));
+        }
+    }
+}