LeetCode 11: Container With Most Water

最新推荐文章于 2020-06-12 18:56:37 发布

claviclelau

最新推荐文章于 2020-06-12 18:56:37 发布

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分类专栏： LeetCode

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本文介绍了一个寻找二维平面上两条直线以形成能容纳最多水的容器的算法。此算法利用双指针技巧，从两端向中间逼近，逐步缩小搜索范围，最终找到最优解。时间复杂度为O(n)。

Difficulty: 3

Frequency: 2

Problem:

Given n non-negative integers a₁, a₂, ..., a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) 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.

Solution:

class Solution {
public:
    int maxArea(vector<int> &height) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (height.size()<2)
            return 0;
            
        int i_start = 0, i_end = height.size() - 1;
        int i_area = (height[i_start]<height[i_end]?height[i_start]:height[i_end])*(i_end - i_start);
        int i_temp_area = 0;
        while(i_start<i_end)
        {
            if (height[i_start]<height[i_end])
                ++i_start;
            else
                --i_end;
            
            i_temp_area = (height[i_start]<height[i_end]?height[i_start]:height[i_end])*(i_end - i_start);
            i_area = i_area>i_temp_area?i_area:i_temp_area;
        }
        return i_area;
    }
};

Notes:

Originally, I misunderstood this problem. This algorithm is borrowed from other.

Time complexity is O(n).

Another thing is that when (height[i_start] == height[i_end]), move either i_start or i_end is OK.