[LeetCode] Container With Most Water

Given n non-negative integers a₁, a₂, ..., a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) 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.

最直观的方法，把每两两之间的容器面积算出来，找出最大的，时间复杂度O（n^2）,不出所料Time Limit Exceeded！这种笨方法如下所示：

class Solution {
public:
    int maxArea(vector<int> &height) {
        int num = height.size();
        int MaxArea = 0;
        if(num<=1)
            return MaxArea;

        for(int i=0;i<num-1;i++){
            for(int j=i+1;j<num;j++)
              MaxArea = max(MaxArea,(j-i)*min(height[i],height[j]));
        }//end for
        return MaxArea;
    }
};

所以要探索简单的方法，遍历所有两两之间的面积中，有很多是无用功，比如S =height[0,len-1];如果height[0]<height[len-1],

那么maxS = max(S,height(1,len-1)),这样就省略了[0,1]、[0,2]....[0,len-2]计算S，

因为此时max([0,1]、[0,2]....[0,len-1]) = height[0,len-1],具体编程如下：

    int maxArea(vector<int> &height){
         int num = height.size();
         int MaxArea = 0;
         if(num<=1)
            return MaxArea;
         int low = 0, high = num-1;
         while(high>low)
         {
         
           MaxArea = max(MaxArea,(high-low)*min(height[high],height[low]));
           if(height[high]>height[low])
               low++;
           else
               high--;
         
         }
        return MaxArea;
    }