Leetcode-container-with-most-water



题目描述

Given n non-negative integers a₁,a₂, ...,a_n, where each represents a point at coordinate (i,a_i).n vertical lines are drawn such that the two endpoints of linei is at (i,a_i) 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.

我采用的就是暴力搜索的方法，复杂度o（n^2），感觉不是太好。

    public static int maxArea(int[] height) {    int max = -1;    int res = 0;        for(int i=0; i<height.length-1; i++){        for(int j=i+1; j<height.length; j++){        if(height[i] >= height[j]){        res = height[j] * (j-i);        }else{        res = height[i] * (j-i);        }        if(res > max)        max = res;        }        }        return max;    }

看了一下别人的做法，用贪心算法解决，感觉是好一些。

    public static int maxArea(int[] height) {        int result=0;        if(height.length<=1) return result;  //无法构成容器        int i=0,j=height.length-1;        while(i<j){            int area = (j-i)*Math.min(height[i],height[j]);            if(area>result)result = area;            if(height[i]>height[j]){                j--;            }else{                i++;            }                     }        return result;    }

合理性解释：当左端线段L小于右端线段R时，我们把L右移，这时舍弃的是L与右端其他线段（R-1, R-2, ...）组成的木桶，这些木桶是没必要判断的，因为这些木桶的容积肯定都没有L和R组成的木桶容积大。