LeetCode - Container With Most Water

题目：

Given n non-negative integers a₁, a₂, ..., a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) 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.

思路：

1）两个for循环，遍历所有情况，时间O(n^2). 超时了.

package area;public class ContainerWithMostWater {public int maxArea(int[] height) {int n = height.length;int maxArea = 0;for (int i = 0; i < n - 1; ++i) {for (int j = i + 1; j < n; ++i) {int area = Math.min(height[i], height[j]) * (j - i);if (maxArea < area)maxArea = area;}}return maxArea;}public static void main(String[] args) {// TODO Auto-generated method stub
}}

 

2）用两边夹的方式解决。设A_ij为坐标(a_i, a_j)所圈定的区域面积，其中j > i，则A_ij = min(a_i, a_j)*(j - i)。

     如果a_i <= a_j，则对于i < k < j，有A_ik = min(a_i, a_k) * (k - i)。由于(k - i) < (j - i)，min(a_i, a_k) <= min(a_i, a_j) ，所以A_ik < A_ij. 这说明，j没必要往左移，这是只能让i往右移。

     如果a_i > a_j，则对于i < k < j，有A_kj = min(a_k, a_j) * (j - k)。由于(j - k) < (j - i)，min(a_k, a_j) <= min(a_i, a_j) ，所以A_kj < A_ij. 这说明，i没必要往右移，这是只能让j往左移。

package area;public class ContainerWithMostWater {public int maxArea(int[] height) {int n = height.length;int maxArea = 0;int i = 0;int j = n - 1;while (i < j) {int area = Math.min(height[i], height[j]) * (j - i);if (height[i] <= height[j]) ++i;else --j;            if (area > maxArea)maxArea = area;}return maxArea;}public static void main(String[] args) {// TODO Auto-generated method stub
}}