【leetcode】11. Container With Most Water

题目描述：

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.

解题思路：

先找距离最长的两个位置看能装下多少水，随后缩小两边的距离，因为距离减少，所以要先使得所能容纳的水变多，只能提高形成的“桶”的高度。因为高度与最低的边界一致，所以缩小边界时要从值小的那一边找，知道找到大于“桶高”的值，说明桶的高度可以得到提升，计算新桶所能容纳的体积并与最大值比较。以此类推，知道两边不能在缩小位置。

具体代码：

public class Solution {
   public static int maxArea(int[] height) {
        if(height==null||height.length<=0)
            return 0;
        if(height.length==2){
            int n=Math.min(height[0], height[1]);
            return n;
        }
        
        
        int from=0;
        int to=height.length-1;
        int min=Math.min(height[from], height[to]);
        int max=(to-from)*min;
        while(from<to){
            if(height[from]<=height[to]){
                from++;
                while(from<to && height[from]<=min){
                    from++;
                }
            }
            else{
                to--;
                while(from<to && height[to]<=min){
                    to--;
                }
            }
            min=Math.min(height[from], height[to]);
            int area=(to-from)*min;
            if(max<area)
                max=area;
        }
        return max;
    }
}