11. Container With Most Water

Author: zhangbo54

src/leetcode/_11_/Main.java

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+package leetcode._11_;
+
+/**
+ * Created by zhangbo54 on 2019-03-01.
+ */
+public class Main {
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        System.out.println(solution.maxArea(new int[]{1, 8, 6, 2, 5, 4, 8, 3, 7}));
+        // Integer.parseInt("1" + Integer.MAX_VALUE);
+    }
+
+    static
+    class Solution {
+        public int maxArea(int[] height) {
+            int i = 0;
+            int j = height.length - 1;
+            int water = 0;
+            while (i < j) {
+                water = Math.max((j - i) * Math.min(height[i], height[j]), water);
+                if (height[i] < height[j]) {
+                    i++;
+                } else {
+                    j--;
+                }
+            }
+            return water;
+        }
+    }
+}

src/leetcode/_11_/md.md

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+### [11\. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)
+
+Difficulty: **Medium**
+
+
+Given _n_ non-negative integers _a<sub style="display: inline;">1</sub>_, _a<sub style="display: inline;">2</sub>_, ..., _a<sub style="display: inline;">n </sub>_, where each represents a point at coordinate (_i_, _a<sub style="display: inline;">i</sub>_). _n_ vertical lines are drawn such that the two endpoints of line _i_ is at (_i_, _a<sub style="display: inline;">i</sub>_) and (_i_, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
+
+**Note: **You may not slant the container and _n_ is at least 2.
+
+![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/17/question_11.jpg)
+
+<small style="display: inline;">The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49\.</small>
+
+**Example:**
+
+```
+Input: [1,8,6,2,5,4,8,3,7]
+Output: 49```
+
+
+#### Solution
+
+Language: **Java**
+
+```java
+class Solution {
+    public int maxArea(int[] height) {
+        int i = 0;
+        int j = height.length - 1;
+        int water = 0;
+        while (i < j) {
+            water = Math.max((j - i) * Math.min(height[i], height[j]), water);
+            if (height[i] < height[j]) {
+                i++;
+            } else {
+                j--;
+            }
+        }
+        return water;
+    }
+}
+```