leetcode_Container With Most Water

标签：dynamic programming   container with most   

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.

Note: You may not slant the container.

思路：

1.先获取两段的垂线和x轴组成的容器可以容纳的水量tempSum，并使得maxSum=tempSum;

2.然后再去除两段较小的垂线段，组成新的容器，获得新的可容纳的水量tempSum，并更新maxSum

3.循环直至start==end

代码：

public int maxArea(int[] height) {
       if(height==null||height.length==1)
            return 0;
       int end=height.length-1,start=0;
       int sum=0;
       while(start<end)
       {
           sum=Math.max(sum,Math.min(height[start],height[end])*(end-start));
           if(height[start]<height[end])
                start++;
            else
                end--;
       }
       return sum;
    }

leetcode_Container With Most Water

标签：dynamic programming   container with most   

原文地址：http://blog.csdn.net/mnmlist/article/details/45815691