leetcode Container With Most Water

首先想到的方法是两层循环，分别计算取最大值，时间复杂度O（n^2）

大数据提示超时

超时的代码

class Solution {
public:
    int maxArea(vector<int> &height) {
        
        int size = height.size();
        int maxWater = 0;
        for(int i = 0; i < size; ++i)
        {
            if(height[i]==0)
                continue;
            for(int j = i+1; j < size; ++j)
            {
                int temp = (j-i)*min(height[i], height[j]);
                if(temp>maxWater)
                    maxWater = temp;
            }
            
        }
        
        return maxWater;
    }
};

参考http://www.cnblogs.com/TenosDoIt/p/3812880.html

Accepted

lass Solution {
public:
    int maxArea(vector<int> &height) {
        
        int left = 0;
        int right = height.size()-1;
        int maxWater = 0;
        
        while(left<right)
        {
            maxWater = max(maxWater, (right-left)*min(height[left], height[right]));
            if(height[left]<height[right])
                left++;
            else
                right--;
            
        }
        return maxWater;
    }
};