Array - 164. Maximum Gap

ppxai

发布于 2020-09-23 17:10:58

2880

发布于 2020-09-23 17:10:58

文章被收录于专栏：皮皮星球皮皮星球

164. Maximum Gap

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.

Return 0 if the array contains less than 2 elements.

Example 1:

Input: [3,6,9,1] Output: 3 Explanation: The sorted form of the array is [1,3,6,9], either (3,6) or (6,9) has the maximum difference 3.

Example 2:

Input: [10] Output: 0 Explanation: The array contains less than 2 elements, therefore return 0.

Note:

You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
Try to solve it in linear time/space.

思路：

找出未排序数组排序后的最大间隔，一般如果是线性的时间复杂度，首先想到桶排序。

代码：

java：

class Solution {

    public int maximumGap(int[] nums) {

         int n = nums.length;
        if(n < 2) return 0;
        int min = nums[0];
        int max = nums[0];
        for(int i = 1;i < n;i++){
            if(min > nums[i]) min = nums[i];
            if(max < nums[i]) max = nums[i];
        }

        int gap = (max-min)/(n-1);
        if(gap == 0) gap++;
        int len = (max-min)/gap+1;
        int [] tmax = new int [len];
        int [] tmin = new int [len];

        for(int i = 0;i < n;i++){
            int index = (nums[i]-min)/gap;
            if(nums[i] > tmax[index]) tmax[index] = nums[i];
            if(tmin[index] == 0 || nums[i] < tmin[index]) tmin[index] = nums[i];
        }
        int myMax = 0;
        for(int i = 0;i < len;i++){
            if(myMax < tmin[i]-min) myMax = tmin[i]-min;
            if(tmax[i] != 0) min = tmax[i];
        }
        return myMax;
    }
}

本文参与腾讯云自媒体分享计划，分享自作者个人站点/博客。

原始发表：2019年06月17日，如有侵权请联系 cloudcommunity@tencent.com 删除

contains