[LeetCode]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.
题意是有个高度数组，就相当于隔板的高度，求数组中任意两隔板间盛水的最大量。隔板间的距离与较低隔板的高度乘积即为盛水的容量。

解题思路

我们来看图分析：



计算桶的容积，应该以最短的那条边来计算，如图中的红线和黑实线所围成的，即min(height[left],height[right])*(right - left)；
我们从left = 0，right = height.length - 1来考虑，此时宽度最大，高度由min(height[left],height[right])来决定，如果height[left] < height[right]，那么如果有最大容积，那么它只能出现在[left+1,right]区间范围内，因为[left,right]长度已是最长，高度最多为height[left],所以为求出最大值则应该left++；
同理如果height[left] > height[right]，则right--；
循环结束的判定为left < right；

代码

public class Solution {
public int maxArea(int[] height) {
int v = 0,water = 0;
int left = 0,right = height.length - 1;

while(left < right){
water = 0;
if(height[left] < height[right]){
water = (right - left)*height[left];
left++;
} else{
water = (right - left)*height[right];
right--;
}
if(water > v){
v = water;
}
}

return v;
}
}