LeetCode 11 Container With Most Water

题目如下：

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.

翻译略去，直接说思路，刚开始我想错了，直接想成了计算题型面积(a[j]-a[i])*(j-i)/2，而且还用了个穷举法。

这里应该计算并比较min(h[i], h[j]) * (j - i)。

选择从两端到中间递归查找，程序如下：

class Solution {
public:
   int maxArea(vector<int> &height)
{
    int ret = findMax(height, 0, height.size()-1, 0);
    return ret;
}
int findMax(vector<int> &h, int i , int j , int area)
{
    int temp = 0;

    if(i == j)
    {
        cout << i << " " << j << endl;
        return area;
    }

    temp = (h[i]>h[j]?h[j]:h[i]) * (j - i) ;
    if(h[i] > h[j])
    {
        j -= 1;
    }
    else
    {
        i += 1;
    }
    if(temp < area)
        findMax(h, i, j , area);
    else
        findMax(h, i, j, temp);
}
};

非递归版本：

class Solution {
public:
   int maxArea(vector<int> &height)
{
    int area = 0;
    int iStart = 0;
    int iEnd = height.size()-1;
    vector<int> h = height;

    while(iStart != iEnd)
    {
        int temp = (h[iStart] > h[iEnd]?h[iEnd]:h[iStart]) * (iEnd - iStart);
        if(temp > area)
        {
            area = temp;
        }
        if(h[iStart] > h[iEnd])
        {
            iEnd -= 1;
        }
        else
        {
            iStart += 1;
        }
    }
    return area;
}
};

加油！