Container With Most Water——LeetCode

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.

 

题目大意：给定一个数组，下标和值分别代表x轴坐标和高度，找出两个元素使得他们之间和x轴围成的范围最大（由短的那个决定高）。

解题思路：一开始想暴力，后来想DP，应该都是O(N^2)的，都过不了，后来看了下提示Two Pointers，前后两个指针，与2Sum的思想类似。

看的比较好的一个解释：

Idea / Proof:

1. The widest container (using first and last line) is a good candidate, because of its width. Its water level is the height of the smaller one of first and last line.
2. All other containers are less wide and thus would need a higher water level in order to hold more water.
3. The smaller one of first and last line doesn't support a higher water level and can thus be safely removed from further consideration.

    public int maxArea(int[] height) {

           if(height == null || height.length<=1){

               return 0;

           } 

           int area=Integer.MIN_VALUE,i=0,j=height.length-1;

           while(i < j){

               area = Math.max(area,(j-i)*Math.min(height[i],height[j]));

               if (height[i]<height[j]) {

                   i++;

               }else{

                   j--;

               }

           }

           return area;

    }