LeetCode: Container With Most Water

11. Container With Most Water

 

Total Accepted: 71232 Total Submissions: 208739 Difficulty: Medium

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.

        暴力解法：遍历所有可能的情况，计算面积，返回最大值，时间复杂度为O(n^2)。很不幸，该方法目前看答案是对的，但是leetcode中提交显示超时。。。

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

     容器面积为h[i]和h[j]中较短的边长乘以i和j间的距离，所以最终的h[maxi]和h[maxj]的左右没有比其自身更高的边。所以我们从两头向中间靠拢，先从较短的边进行靠拢，找到比其更高的边再进行判断。

class Solution {
public:
    int maxArea(vector<int>& height) {
        int maxarea = 0;
        int i = 0, j = height.size() - 1;
        while (i<j)
        {
            int hei = height[i] > height[j] ? height[ j] : height[i];
            if (maxarea < hei*abs(i - j))
            {
                maxarea = hei*abs(i - j);
                
            }
            if (height[i] < height[j])
            {
                int k = i;
                
                while (height[k]<=height[i]&&k<j)
                {
                    k++;
                }
                i = k;
            }
            else
            {
                int h = j;
                while (height[h] <= height[j] && i<h)
                {
                    h--;
                }
                j = h;
            }
        }
        return maxarea;
    }
};

 

附：vector中size和capacity,"[]"和at()的区别，

     size指目前容器中实际有多少元素，对应的resize(size_type)会在容器尾添加或删除一些元素，来调整容器中实际的内容，使容器达到指定的大小。capacity指最少要多少元素才会使其容量重新分配，对应reserve(size_type new_size)会这置这个capacity值，使它不小于所指定的new_size。

     at()提供了越界检查的功能，使用起来更安全，同时比起operator[]来代价也更大。

参考链接：http://blog.163.com/jxguo_05/blog/static/719401002010102523831645/

http://blog.csdn.net/a83610312/article/details/8548519