给你n个任意整数,求排序后相邻两个数之间的最大差值,这里n可能有10^5,整数为任意32位整型。要求求解算法的时间复杂度为O(n)。
思路描述参考:求无序数组排序后相邻俩数最大差值
代码C++描述如下:
#include <iostream>
using namespace std;
int max(int x, int y)
{
int z = x > y ? x : y;
return z;
}
int min(int x, int y)
{
int z = x > y ? y : x;
return z;
}
int bucket(long num, long len, long min, long max)
{
return (int)((num - min) * len / (max - min));
}
int maxGap(int nums[], int length)
{
if(nums == NULL || length < 2) {
return 0;
}
int minValue = 32768;
int maxValue = -32767;
for(int i = 0; i < length; i++) {
minValue = min(minValue, nums[i]);
maxValue = max(maxValue, nums[i]);
}
if(minValue == maxValue) {
return 0;
}
int hasNum[length + 1] = {0};
int maxs[length + 1] = {0};
int mins[length + 1] = {0};
int bid = 0;
for(int i = 0; i < length; i++) {
bid = bucket(nums[i], length, minValue, maxValue);
mins[bid] = hasNum[bid] ? min(mins[bid], nums[i]) : nums[i];
maxs[bid] = hasNum[bid] ? max(maxs[bid], nums[i]) : nums[i];
hasNum[bid] = true;
}
int res = 0;
int lastMax = maxs[0];
int i = 1;
for(; i <= length; i++) {
if(hasNum[i]) {
res = max(res, mins[i] - lastMax);
lastMax = maxs[i];
}
}
return res;
}
int main(){
int nums[]={0,89,34,22};
int length = sizeof (nums)/sizeof(nums[0]);
int res = maxGap(nums,length);
printf("res is %d\n",res);
}