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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.
Credits:
Special thanks to @porker2008 for adding this problem and creating all test cases.
// 用桶排序
// 算出相邻两个桶之间的最大差值
// 如果是平均分布，则桶的数目和元素的数目相同时，其排序的时间复杂度是0(n)
// 我们假设桶的个数和元素的数目相同，若是平均分布，则每个桶里有一个数，而若某个桶里有两个以上的桶时，这时必有至少一个是空桶，那么最大间隔可能就落在空桶的相邻两个桶存储的数之间，最大间隔不会落在同一个桶的数里，因此我们不需要对每个桶再排一次序，只需要记录同一个桶的最大值和最小值，算出前一个有最大值的桶和后一个有最小值的桶之差，则可能是最大间隔
//步骤：1.算好用的桶的个数，用最大元素和最小元素算出平均间隔，记录在平均间隔上的最大值和最小值，
// 2. 再算出前一个间隔里的最大值和后一个间隔里的最小值之差，取最大的一个，
#include <stdio.h> #include <stdlib.h> #include <iostream> #include <vector> #include <queue> #include <stack> #include <math.h> #include <limits.h> using namespace std; void printArray(int *array, int size) { for(int i = 0; i < size; i++) cout << array[i]<< "\t" ; cout << endl; } void printVector(vector<int> array ) { for(int i = 0; i <array.size(); i++) cout << array[i]<< "\t" ; cout << endl; } class Solution { public: int maximumGap(vector<int>& nums) { int n = nums.size(); if(n == 0 || n == 1) return 0; int maxVal = nums[0]; int minVal = nums[0]; for(int i = 1; i < n; i++) { maxVal = max(maxVal, nums[i]); minVal = min(minVal, nums[i]); } // n nums has (n-1) the gap ; // average gap = (maxVal - minVal) / (n-1); // so the minimum of max gap is ceilOf(average gap); int minOfMaxGap = ceil((double)(maxVal - minVal) /(n - 1)); // bucket number, the last bucket may contains the maxVal // Èç¹ûÄÜÕû³ý£¬ÄÇômaxValÔÙ×îºóÒ»¸öbucket, // Èç¹û²»ÄÜÕû³ý£¬ÄÇômaxValÔÚµ¹ÊýµÚ¶þ¸öbucket int bucketCnt = ceil((double)(maxVal - minVal)/minOfMaxGap) + 1; // for numbers in a same bucket, its max gap is minOfMaxGap, // So the maxGap occur at different buckets, including emtpy buckets // we record maxVal and minVal in every bucket vector<int> maxVec(bucketCnt, INT_MIN); vector<int> minVec(bucketCnt, INT_MAX); int idx = 0; for(int i = 0; i < n; i ++) { idx = (nums[i] - minVal)/minOfMaxGap; if(maxVec[idx] < nums[i]) maxVec[idx] = nums[i]; if(minVec[idx] > nums[i]) minVec[idx] = nums[i]; } //cout << "minOfMaxGap\t" << minOfMaxGap << endl; //cout << "bucketCnt\t"<< bucketCnt<< endl; //cout << "maxVec\t"; //printVector(maxVec); //cout << "minVec\t"; //printVector(minVec); int maxGap = minOfMaxGap; int preMax = maxVec[0]; for(int i = 1; i < bucketCnt; i++) { if(minVec[i] == INT_MAX)// empty bucket continue; if(minVec[i] - preMax > maxGap) { maxGap = minVec[i] - preMax; } preMax = maxVec[i]; } return maxGap; } }; int main() { vector<int> nums; nums.push_back(3); nums.push_back(6); nums.push_back(9); nums.push_back(1); Solution sl; cout << sl.maximumGap(nums) <<endl; return 0; }
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原文地址：http://www.cnblogs.com/diegodu/p/4643559.html
