leetcode题解-11. Container With Most Water

题目：

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 and n is at least 2.

本题是为了寻找最大容积的两条直线。需要衡量的点主要是高度和宽度之间的权衡。首先我们想到的最简单的思路应该是使用暴力搜索法，使用两次遍历的方法分别对每个数搜索其最大容积。代码入下：

    public int maxArea(int[] height) {        int max = 0;        for(int i=0; i<height.length-1; i++){            int tmp = height[i];            for(int j=i+1; j<height.length; j++){                max = Math.max(max, (j-i)*Math.min(tmp, height[j]));            }        }        return max;    }

但是这种方法会产生TLE，时间超出限制。所以我们相到下面的方法，分别从左右两边开始搜索，用max记录最大值，然后不断遍历。每次遍历的依据是看左右两边的值那个大，将小的移向下一位置。代码入下：

    public int maxArea1(int[] height){        int max = 0, left=0, right=height.length-1;        while(left < right){            if(height[left] < height[right]){                max = Math.max(max, (right-left)*height[left]);                left ++;            }else{                max = Math.max(max, (right-left)*height[right]);                right --;            }        }        return max;    }

上面这种方法客击败88。5%的用户，但是我们很显然还可以进行改进，就是每次都移动多次，这样可以省去计算最大值的时间，击败99。2%的用户。代码入下：

    public int maxArea2(int[] height) {        int lo = 0;        int hi = height.length - 1;        int max = 0;        while(lo < hi) {            int min = Math.min(height[lo], height[hi]);            max = Math.max(max, min * (hi - lo));            while(lo <= hi && height[lo] <= min) lo++;            while(lo <= hi && height[hi] <= min) hi--;        }        return max;    }