[Leetcode] 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.

Solution: 

这道题起初我理解错了题意，以为是要找到选定数组中最小的来乘以（r-l）以此为基础来求组成的面积的最大值，其实这样是错的。

看了大神的解答后，这道题就是只看选中数组左边和右边的值，选择其中最小的那个来乘以（r-l），以此为基础来求组成的面积的最大值。

Solution : 2 pointers分别指向array的开头和结尾

public class Solution {
    public int maxArea(int[] height) {
        int N=height.length;
        if(N<2)
            return 0;
     //   System.out.println("N==="+N);
        int maxarea=0;
        int l=0,r=N-1;
        while(l<r){
            if(height[l]<height[r]){
                int temp=(r-l)*height[l];
                ++l;
                maxarea=Math.max(temp, maxarea);
            }else{
                int temp=(r-l)*height[r];
                --r;
                maxarea=Math.max(temp, maxarea);
            }
        }
        return maxarea;
    }
}

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;
    }
};