19.1.25 [LeetCode11]Container With Most Water

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.

 

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

 

Example:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

题解

题意：一组数组a[]，求两个索引x,y，使得 (y-x)*min(a[x],a[y]) 最大

一开始的方法，暴力：

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

后来看了下题解发现可以这样：设置两个头尾指针向中间移动取值更新答案

class Solution {
public:
    int maxArea(vector<int>& height) {
        int s = 0, e = height.size() - 1, ans = 0;
        while (s != e) {
            ans = max(ans, min(height[s], height[e])*(e - s));
            if (height[s] < height[e])
                s++;
            else e--;
        }
        return ans;
    }
};

（每次改变较短边的索引，保证这种移动方式比改变较长边得到的答案一定好，而每一个最优解都能从初始状态通过移动两端指针得到，这就证明了正确性）

然而还是慢！下面的是翻了别人的帖子看到的题解，就是个小优化，快了4ms吧，但排名上升了很多

class Solution {
public:
    int maxArea(vector<int>& height) {
        int s = 0, e = height.size() - 1, ans = 0;
        while (s != e) {
            int h = min(height[s], height[e]);
            ans = max(ans, h*(e - s));
            while (height[s] <= h && s < e)
                s++;
            while (height[e] <= h && s < e)
                e--;
        }
        return ans;
    }
};