LeetCode || Container With Most Water

Container With Most Water

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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1：找最大面积问题，取决于较小的高度，area=（j-i）*min（height[i], height[j]）。思路一是两层遍历，计算每一个可能的组合，取最大值，代码如下。

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

思路2：思路1的复杂度为O(n^2)，会超时；思路2利用找最大和的思想，用两个指针 i、j 从数组的两头开始向中间走，计算面积值，然后移动其中height较小的那个指针，直至 i>=j 结束，复杂度O(N)，代码如下。

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