【11】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.
从两边向中间搜(这样可以保证宽度是在单调减少的,利于确定高度的变化)
那么初始设i=0,j=n-1
就是一根柱子在最左边,一根柱子在最右边,然后我们舍弃那根较短的柱子,并从他的位置开始向中间寻找,直到找到一根比之前长的柱子,判断一下此时的盛水量是不是比之前多了,直到两根柱子相遇(即i==j)。为什么这里要舍弃那根较短的柱子呢,因为如果舍弃较长的那根,此时高度不会有所增加,短板依然是之前较短的那根柱子,而宽度在减少,所以肯定找不到比之前装水更多的位置
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 nH=height[l]; while(l<r&&height[l]<=nH)l++; } else{ int nH=height[r]; while(r>l&&height[r]<=nH)r--; } } return res;}
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