算法系列——Container With Most Water

题目描述

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 and n is at least 2.

解题思路

用两个指针从两端开始向中间靠拢，如果左端线段短于右端，那么左端右移，反之右端左移，知道左右两端移到中间重合，记录这个过程中每一次组成木桶的容积，返回其中最大的。

原因：当左端线段L小于右端线段R时，我们把L右移，这时舍弃的是L与右端其他线段（R-1, R-2, …）组成的木桶，这些木桶是没必要判断的，因为这些木桶的容积肯定都没有L和R组成的木桶容积大。

程序实现

public class Solution {    public int maxArea(int[] height) {        if(height==null||height.length<2)            return 0;        int left=0;        int right=height.length-1;        int maxVal=0;        while(left<right){            int curVal=(right-left)*Math.min(height[left],height[right]);            if(curVal>maxVal)                maxVal=curVal;             if (height[left] < height[right])                  left++;              else                right--;          }        return maxVal;    }}