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Given an unsorted array, find the maximum difference between the successive elements in its sorted form.
Try to solve it in linear time/space.
Return 0 if the array contains less than 2 elements.
You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
class Solution{ public: int maximumGap(vector<int>& nums){ int n = nums.size(); if( n < 2) return 0; int maxVal = *max_element(nums.begin(),nums.end()); int minVal = *min_element(nums.begin(),nums.end()); if( maxVal == minVal) return 0; int gap = int((maxVal-minVal)/n+1); vector<int> maxofBucket(n,INT_MIN); vector<int> minofBucket(n,INT_MAX); for(int i=0;i<n;i++){ int bucketId = int((nums[i]-minVal)/gap); maxofBucket[bucketId] = max(nums[i],maxofBucket[bucketId]); minofBucket[bucketId] = min(nums[i],minofBucket[bucketId]); } int preMax = maxofBucket[0]; int res = INT_MIN; for(int i=1;i<n;i++){ if(minofBucket[i] != INT_MAX){ res = max(res,minofBucket[i]-preMax); preMax = maxofBucket[i]; } } return res; } };
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原文地址：http://www.cnblogs.com/wxquare/p/5951823.html
