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 and n is at least 2.
下面以例子: [4,6,2,6,7,11,2] 来讲解。
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;
直观的解释是:容积即面积,它受长和高的影响,当长度减小时候,高必须增长才有可能提升面积,所以我们从长度最长时开始递减,然后寻找更高的线来更新候补;
class Solution {public: int maxArea(vector<int>& height) { size_t i = 0; size_t j = height.size()-1; int mastWater = 0; while(i < j) { int newWater = (j-i)*min(height[i], height[j]); mastWater = max( mastWater,newWater ); if(height[i] < height[j]) { int x = i; while(x < j && height[x] <= height[i]) x++; i = x; } else { int x = j; while(i < x && height[x] <=height[j]) x--; j = x; } } return mastWater; }};
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