【leetcode】Container With Most Water（middle）

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.

思路：

肯定是要确定水槽的左右边界位置，需要两个变量l1,l2 分别从最左边和最右边收缩。收缩规则是，比较矮的那一边向中间靠拢。更新水量。

我自己的代码：

int maxArea(vector<int> &height) {
        int ans = (height.size() - 1) * min(height[0], height.back()) ;
        int l1, l2, h1, h2;
        h1 = 0; h2 = height.size() - 1;
        l1 = h1; l2 = h2;
        while(l1 < l2)
        {
            if(height[l1] > height[h1])
            {
                h1 = l1;
                int tmp = (l2 - l1) * min(height[l2], height[l1]);
                ans = (tmp > ans) ? tmp : ans;
            }
            if(height[l2] > height[h2])
            {
                h2 = l2;
                int tmp = (l2 - l1) * min(height[l2], height[l1]);
                ans = (tmp > ans) ? tmp : ans;
            }
            if(height[l1] < height[l2])
            {
                l1++;
            }
            else if(height[l1] > height[l2])
            {
                l2--;
            }
            else
            {
                l1++; l2--;
            }
        }
        return ans;
    }

大神的代码就简洁很多：

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