11. Container With Most Water

Given n non-negative integers a1a2, ..., an , where each represents a point at coordinate (iai). n vertical lines are drawn such that the two endpoints of line i is at (iai) 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.

 

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

 

Example:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49
class Solution {
    public int maxArea(int[] height) {
        int res = 0;
        for(int i = 0; i < height.length; i++){
            for(int j = i + 1; j < height.length; j++){
                int min = Math.min(height[i], height[j]);
                res = Math.max((j - i) * min, res);
            }
        }
        return res;
    }
}

花里胡哨的,先给你来个brute force。

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

双指针法,从左到右扫描一次,每次更新max(比较max和当前value(以左右低的为基准))。

posted @ 2019-09-26 11:31  Schwifty  阅读(118)  评论(0编辑  收藏  举报