[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.

题意：假设有一个x,y的二维坐标，我们定义area(i,j) = min(xi,xj) * |i - j, 找出最大的area。
思路：用两个指针l,r分别指向数组两边，然后计算当前area，接着比较xl,xr大小，小的那个的下标变化。知道l>=r|。 这是由于当前的矩形宽度是最大的，那么接下来的矩形想想面积比当前的面积大，就只能够增加矩形的高。算法复杂度为O(N)
代码：

    public int maxArea(int[] height) {//O(N)
        int max = 0;
        int curr  = 0;
        int l = 0, r = height.length - 1;
        while (l < r){
            curr = Math.min(height[l],height[r]) * (r - l);
            max = Math.max(max, curr);
            if(height[l] < height[r]){
                l ++;
            }else if(height[l] > height[r]){
                r  --;
            }else {
                l ++;
                r  --;
            }
        }
        return max;
    }

[LeetCode]Container With Most Water

原文：http://blog.csdn.net/youmengjiuzhuiba/article/details/45176327