[leetcode] Container With Most Water
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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.
原文链接:http://blog.csdn.net/a83610312/article/details/8548519
思路:
1.首先假设我们找到能取最大容积的纵线为 i , j (假定i<j),那么得到的最大容积 C = min( ai , aj ) * ( j- i) ;
2.下面我们看这么一条性质:
①: 在 j 的右端没有一条线会比它高! 假设存在 k |( j<k && ak > aj) ,那么 由 ak> aj,所以 min( ai,aj, ak) =min(ai,aj) ,所以由i, k构成的容器的容积C' = min(ai,aj ) * ( k-i) > C,与C是最值矛盾,所以得证j的后边不会有比它还高的线;
②:同理,在i的左边也不会有比它高的线;
这说明什么呢?如果我们目前得到的候选: 设为 x, y两条线(x< y),那么能够得到比它更大容积的新的两条边必然在 [x,y]区间内并且 ax' > =ax , ay'>= ay;
3.所以我们从两头向中间靠拢,同时更新候选值;在收缩区间的时候优先从 x, y中较小的边开始收缩;
代码:
class Solution {public: int maxArea(vector<int> &height) { int res=0; int n=height.size(); int l=0,r=n-1; while(l<r){ res=max(res,min(height[l],height[r])*(r-l)); if(height[l]<height[r]){ int k=l; while(k<r&&height[k]<=height[l]) k++; l=k; } else{ int k=r; while(k>l&&height[k]<=height[r]) k--; r=k; } } return res; }};
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