11. Container With Most Water

发表于 | 分类于

每日一题 2019 - 03 - 03

题目：

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.

img

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:

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

解法：

这个题让求出最大的盛水量，也即找出最大的矩形面积，所以可以直接用暴力的方法求解，也即使用两个指针一前一后进行收缩变换，同时对已经求出的最大盛水量进行比较，更新最大值，即可完成求解，但是需要注意一点，在对指针进行收缩变换的时候，要注意每次变动的一定是小的那一边，这样能够保证每次都用最大的边去作为边界。

代码：

1
2
3
4
5
6
7
8
9
10
11
class Solution {
public:
    int maxArea(vector<int>& height) {
        int res = 0, i = 0, j = height.size() - 1;
        while (i < j) {
            res = max(res, min(height[i], height[j]) * (j - i));
            height[i] < height[j] ? ++i : --j;
        }
        return res;
    }
};
0%
undefined