[LeetCode]Container With Most Water
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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; }}
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