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164. Maximum Gap

难度:Hard

刷题内容

原题连接

  • https://leetcode.com/problems/maximum-gap/

内容描述

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.

˼·1 **- ʱ�临�Ӷ�: O(NlgN)*- �ռ临�Ӷ�: O(1)***

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class Solution {
public:
    int maximumGap(vector<int>& nums) {
        if(nums.size() < 2)
            return 0;
        sort(nums.begin(),nums.end());
        int ans = INT_MIN;
        for(int i = 1;i < nums.size();++i)
            ans = max(ans,nums[i] - nums[i - 1]);
        return ans;
    }
};

˼·2 **- ʱ�临�Ӷ�: O(N)*- �ռ临�Ӷ�: O(N)***

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class Solution {
public:
    int maximumGap(vector<int>& nums) {
        if(nums.size() < 2)
            return 0;
        int res[nums.size()];
        int t[nums.size()][10];
        for(int i = 0;i < nums.size();++i)
        {
            res[i] = i;
            for(int j = 0;j < 10;++j)
                t[i][j] = 0;
            int temp = nums[i],j = 9;
            while(temp)
            {
                t[i][j--] = temp % 10;
                temp /= 10;
            }
        }
        for(int i = 9;i >= 0;--i)
        {
            int bucket[nums.size()],count1[10];
            memset(count1,0,sizeof(count1));
            for(int j = 0;j < nums.size();++j)
                count1[t[res[j]][i]]++;
            for(int j = 1;j < 10;++j)
                count1[j] += count1[j - 1];
            for(int j = nums.size() - 1;j >= 0;--j)
                bucket[--count1[t[res[j]][i]]] = res[j];
            for(int j = 0;j < nums.size();++j)
                res[j] = bucket[j];
        }
        int ans INT_MIN;
        for(int i = 1;i < nums.size();++i)
            ans = max(ans,nums[res[i]] - nums[res[i - 1]]);
        return ans;
    }
};


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