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题目：
Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) 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Aij为坐标(ai, aj)所圈定的区域面积，其中j > i，则Aij = min(ai, aj)*(j - i)。
如果ai <= aj，则对于k < j，有Aik = min(ai, ak) * (k - i)。由于(k - i) < (j - i)，min(ai, ak) <= min(ai, aj) ，所以Aik < Aij. 这说明，j没必要往左移，这是只能让i往右移。
如果ai > aj，则对于k < i，有Aik = min(ak, aj) * (j - k)。由于(j - k) < (j - i)，min(ak, aj) <= min(ai, aj) ，所以Aik < Aij. 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 } }
LeetCode - Container With Most Water
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原文地址：http://www.cnblogs.com/shuaiwhu/p/5047844.html
