最大间距

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问题描述:给定n个实数x1,x2,...,xn,求这n(n>=2)个实数在实轴上相邻2个数之间的最大差值,要求设计线性的时间算法

#include <stdio.h>#include <stdlib.h>void MaxGap(int n,double *num){int i,index;double *low = (double *)malloc(sizeof(double)*n); //每个鸽舍的最小数据double *high = (double *)malloc(sizeof(double)*n); //每个鸽舍的最大数据 int *flag = (int *)malloc(sizeof(int)*n); //标记鸽舍是否为空 double avr_gap,minx,maxx,max_gap;//确定间距 minx = maxx = num[0];for(i=1;i<n;i++){if(num[i] < minx)minx = num[i];if(num[i] > maxx)maxx = num[i];}avr_gap = (maxx - minx)/(n-1);if(minx == maxx)printf("0\n");else if(n == 2){printf("%lf\n",maxx - minx);}else{//入鸽,只确定最大和最小值//memset(flag,0,sizeof(flag));//用memset对malloc开辟的数组 初始化只能修改flag[0]的值 for(i=0;i<n;i++){//注意:首先初始化low[i] = maxx;high[i] = minx;flag[i] = 0;}//将除minx和maxx之外的(n-2)个数放在(n-1)个鸽舍中, //由抽屉原理可知至少有一个鸽舍是空的,又因为每个鸽舍大小相同,所以最大间隙不会在同一个鸽舍中出现,//一定是某个鸽舍的上界和其后某个鸽舍的下界之差 //而将minx和maxx放入鸽舍(0鸽舍和(n-1)鸽舍)仍然有这个结论 for(i=0;i<n;i++){  index = (int)((num[i]-minx)/avr_gap);//鸽舍左闭右开(除最后一个鸽舍) if(!flag[index])flag[index] = 1; if(low[index] > num[i]) low[index] = num[i]; if(high[index] < num[i]) high[index] = num[i];}//MaxGap = max(Minj-Maxi)for(i=1;i<n;i++)if(flag[i])break;max_gap = low[i] - high[0];index = i;for(++i;i<n;i++){if(flag[i]){if(low[i] - high[index] > max_gap)max_gap = low[i] - high[index];index = i;}}printf("%lf\n",max_gap);}}int main(){int n,i;double *num;printf("n:");scanf("%d",&n);num = (double *)malloc(sizeof(double)*n); for(i=0;i<n;i++)scanf("%lf",&num[i]);//用鸽舍法求最大间隙问题(用n-2个等间距点分割区间[minx,maxx],产生n-1个段)MaxGap(n,num);return 0;}

样例:
5
2.3 3.1 7.5 1.5 6.3---->3.2
5
2.3 3.1 10.0 1.5 6.3---->3.7
10
2.5 7.2 6.3 4.9 10.6 5.3 4.8 3.0 10.5 5.7
排序:2.5 3.0 4.8 4.9 5.3 5.7 6.3 7.2 10.5 10.6 3.3----->3.3

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