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.
Solution:
1 public class Solution { 2 public int maximumGap(int[] num) { 3 int len = num.length; 4 if (len<2) return 0; 5 if (len==2) return num[1]-num[0]; 6 7 int numMax = num[0], numMin = num[0]; 8 for (int i:num){ 9 numMax = Math.max(numMax,i); 10 numMin = Math.min(numMin,i); 11 } 12 double temp = (double) numMax - numMin; 13 int gap = (int) Math.ceil(temp/(len-1)); 14 15 int[] bucketMin = new int[len-1]; 16 int[] bucketMax = new int[len-1]; 17 Arrays.fill(bucketMin, Integer.MAX_VALUE); 18 Arrays.fill(bucketMax, Integer.MIN_VALUE); 19 for (int i:num){ 20 int index = (i-numMin)/gap; 21 if (i==numMax) index = len-2; 22 if (i>bucketMax[index]) bucketMax[index] = i; 23 if (i<bucketMin[index]) bucketMin[index] = i; 24 } 25 26 int res = Integer.MIN_VALUE; 27 int pre = bucketMax[0]; 28 for (int i=1;i<len-1;i++) 29 if (bucketMin[i] == Integer.MAX_VALUE) continue; 30 else { 31 if (res<(bucketMin[i]-pre)) 32 res = bucketMin[i]-pre; 33 pre = bucketMax[i]; 34 } 35 36 return res; 37 } 38 }
原文：http://www.cnblogs.com/lishiblog/p/4200261.html