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.
给定一个未排序的数组,然后找出排序之后的数组中,相邻数字的最大差。
1、桶排序
public class Solution {
public int maximumGap(int[] nums) {
int len = nums.length;
if (len < 2){
return 0;
}
int max = nums[0];
int min = nums[0];
for (int num : nums){
if (max < num){
max = num;
} else if ( min > num){
min = num;
}
}
int gap = (max-min)/(len-1);
if( gap == 0){
gap = 1;
}
int size = (max - min) / gap + 1;
int[] gapMax = new int[size];
int[] gapMin = new int[size];
for (int num : nums){
int pos = (num - min)/gap;
if (gapMax[pos] < num){
gapMax[pos] = num;
}
if (gapMin[pos] == 0 || gapMin[pos] > num){
gapMin[pos] = num;
}
}
int start = min;
int end = gapMax[0];
int result = end - start;
for (int i = 0; i < size - 1; i++){
start = gapMax[i] == 0 ? start : gapMax[i];
end = gapMin[i+1];
if (result < (end - start)){
result = end - start;
}
}
if (gapMax[size - 1] == 0 && end - start > result){
result = end - start;
} else if (gapMax[size - 1] != 0 && end - gapMax[size - 1] > result){
result = end - gapMax[size - 1];
}
return result;
}
}