Leetcode 11. Container With Most Water
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1.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.
给你一个数组,数组的下标是代表线在x轴上的位置,值代表此条线的高度。要你找到两条线,使这两条线和x轴所围成的容器容量最大(可以想象成一个桶,能接多少的水)
2.思路
我们知道,桶的容量(capabality)=桶的高度(height)*桶的宽度(weight)。
我们先计算nums[0] 和 nums[n-1]所组成的桶的容量,此时桶的宽度是最长的,如果还有容量比它大的桶的话,那么只有高度比它高才有可能,因为宽度是不断递减的。因此我们设立left和right,分别代表桶的左右两边,如果nums[left] < nums[right],那么从左向右寻找比nums[left]高的边;同理,如果nums[left] > nums[right],那么从右向左寻找比nums[right]高的边。
class Solution {public: int maxArea(vector<int>& height) { if(height.size() <= 1) return 0; int left = 0,right = height.size()-1; int ans=0; while(left < right){ int capability = min(height[left],height[right])*(right-left); ans = max(ans,capability); if(height[left] > height[right]) right--; else left++; } return ans; }};
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