LeetCode-Container With Most Water-zz

先上代码。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class Solution {
public:
    int maxArea(vector<int> &height) {
        if (height.empty()) return 0;
        int i = 0, j = height.size() - 1;
        int max = INT_MIN;
        int area;
        while (i < j)
        {
            area = min(height[i],height[j])*(j-i);
            if (area>max) max = area;
            if (height[i] < height[j]) i++;//谁是短板，谁就该移动
            else j--;
        }
        return max;
    }
};

int main()
{
    Solution s;
    int array[] = {2,5,3,1,6};
    vector<int> h(array,array+5);
    cout << s.maxArea(h)<<endl;
    return 0;
}

http://blog.unieagle.net/2012/09/16/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Acontainer-with-most-water/

最笨的方法是暴力破解，加上一点预判。时间是O(n^2)

看了nandawys的评论，找到了O(n)方法，思路是从两头到中间扫描，设i,j分别指向height数组的首尾。
那么当前的area是min(height[i],height[j]) * (j-i)。
当height[i] < height[j]的时候，我们把i往后移，否则把j往前移，直到两者相遇。

这个正确性如何证明呢？
代码里面的注释说得比较清楚了，即每一步操作都能保证当前位置能取得的最大面积已经记录过了，而最开始初始化的时候最大面积记录过，所以有点类似于数学归纳法，证明这个算法是正确的。

Container With Most Water
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.

代码
600ms过大集合

class Solution {
public:
    int maxArea(vector<int> &height) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int max = 0;
        for( int i = 0 ; i < height.size(); ++i)
        {
            int hi = height[i];
            if(hi == 0)
                continue;
            int minPosibleIndex = max / hi + i;
            for(int j = height.size() - 1; j > i && j >= minPosibleIndex; --j)
            {
                int hj = height[j];
                int area = min(hi,hj) * (j - i);
                if (area > max)
                    max = area;
            }
        }
        return max;
    }
};

Code rewrite at 2013-1-4,O(n)

class Solution {
public:
    int maxArea(vector<int> &height) {
        if (height.size() < 2) return 0;
        int i = 0, j = height.size() - 1;
        int maxarea = 0;
        while(i < j) {
            int area = 0;
            if(height[i] < height[j]) {
                area = height[i] * (j-i);
                //Since i is lower than j, 
                //so there will be no jj < j that make the area from i,jj 
                //is greater than area from i,j
                //so the maximum area that can benefit from i is already recorded.
                //thus, we move i forward.
                //因为i是短板，所以如果无论j往前移动到什么位置，都不可能产生比area更大的面积
                //换句话所，i能形成的最大面积已经找到了，所以可以将i向前移。
                ++i;
            } else {
                area = height[j] * (j-i);
                //the same reason as above
                //同理
                --j;
            }
            if(maxarea < area) maxarea = area;
        }
        return maxarea;
    }
};