LeetCode 11: Container With Most Water

Container With Most Water

Given n non-negative integers a₁, a₂, …, a_n, 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.

解题思路

根据题意，需要求出 max(|i - j| * min(a_i, a_j))。

思路一：穷举法，时间复杂度为O(n²)。

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

思路二：使用两个指针 left = 0 和 right = n - 1 分别指向数组的头和尾。重复如下步骤直到 left >= right：首先计算(right - left) * min(a_left, a_rigth)，并更新结果；然后如果 a[left] < a[right], 则left++,否则right–。

这是一个贪心的策略，每次取两边围栏最矮的一个推进，希望获取更多的水。时间复杂度为O(n)。

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