Maximum Gap

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.

利用桶排序来缩减分隔距离，减少内存消耗

1. int maximumGap(vector<int> &num) {
2. unsigned int nSize = num.size();
3. if(nSize < 2) return 0;
5. int maxNum = *max_element(num.begin(),num.end());
6. int minNum = *min_element(num.begin(),num.end());
7. int gap = (int)ceil((double)(maxNum - minNum)/(nSize-1));
8. int bucketNum = (int)ceil((double)(maxNum-minNum)/gap);
9. vector<int> minBucket(bucketNum,INT_MAX);
10. vector<int> maxBucket(bucketNum,INT_MIN);
11. for(int i=0;i<nSize;i++) {
12. if(num[i] == maxNum || num[i] == minNum) {
13. continue;
14. }
15. int bucketId = (num[i]-minNum)/gap;
16. minBucket[bucketId] = min(minBucket[bucketId],num[i]);
17. maxBucket[bucketId] = max(maxBucket[bucketId],num[i]);
18. }
19. int maxGap = INT_MIN;
20. int previous = minNum;
21. for(int i=0;i<bucketNum;i++) {
22. if(minBucket[i] == INT_MAX && maxBucket[i] == INT_MIN) {
23. continue;
24. }
25. maxGap = max(maxGap,minBucket[i]-previous);
26. previous = maxBucket[i];
27. }
28. maxGap = max(maxGap,maxNum-previous);
29. return maxGap;
30. }