《编程珠玑》中的一些代码

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位图排序

使用位图对[0..N-1]中不同的整数进行排序

/* bitmap sort -- Sort distinct integers in the range [0..N-1] */#include <stdio.h>#define BITSPERWORD 32#define SHIFT 5#define MASK 0x1F#define N 10000000int a[1 + N/BITSPERWORD];void set(int i) {    a[i>>SHIFT] |=  (1<<(i & MASK)); }void clr(int i) {    a[i>>SHIFT] &= ~(1<<(i & MASK)); }int  test(int i){    return a[i>>SHIFT] & (1<<(i & MASK)); }int main(){int i;for (i = 0; i < N; i++)clr(i);/*Replace above 2 lines with below 3 for word-parallel initint top = 1 + N/BITSPERWORD;for (i = 0; i < top; i++)a[i] = 0;*/while (scanf("%d", &i) != EOF)set(i);for (i = 0; i < N; i++)if (test(i))printf("%d\n", i);return 0;}

从0...n-1中随机选择m个数

/* sortedrand.cpp -- output m sorted random ints in U[0,n) */#include <iostream>#include <set>#include <algorithm>using namespace std;int bigrand(){return RAND_MAX*rand() + rand();}int randint(int l, int u){return l + bigrand() % (u-l+1);}void genknuth(int m, int n){for (int i = 0; i < n; i++)/* select m of remaining n-i */if ((bigrand() % (n-i)) < m){cout << i << "\n";m--;}}void gensets(int m, int n){set<int> S;set<int>::iterator i;while (S.size() < m){int t = bigrand() % n;S.insert(t);}for (i = S.begin(); i != S.end(); ++i)cout << *i << "\n";}void genshuf(int m, int n){int i, j;int *x = new int[n];for (i = 0; i < n; i++)x[i] = i;for (i = 0; i < m; i++){j = randint(i, n-1);int t = x[i]; x[i] = x[j]; x[j] = t;}sort(x, x+m);for (i = 0; i < m; i++)cout << x[i] << "\n";}void genfloyd(int m, int n){set<int> S;set<int>::iterator i;for (int j = n-m; j < n; j++){int t = bigrand() % (j+1);if (S.find(t) == S.end())S.insert(t); // t not in SelseS.insert(j); // t in S}for (i = S.begin(); i != S.end(); ++i)cout << *i << "\n";}int main(int argc, char *argv[]){int m = atoi(argv[1]);int n = atoi(argv[2]);genknuth(m, n);return 0;}

寻找字符串中最长的重复子串

使用后缀数组，字符串保存在c中，后缀数组为a，C代码：

/* longdup.c -- Print longest string duplicated 1 times */#include <stdlib.h>#include <string.h>#include <stdio.h>int pstrcmp(const void *p, const void *q){    return strcmp(*(char **)p, *(char **)q);}int comlen(char *p, char *q){    int i = 0;while (*p && (*p++ == *q++))i++;return i;}#define MAXN 5000000char c[MAXN], *a[MAXN];int main(){    int i, ch, temlen, n = 0, maxi, maxlen = -1;    while ((ch = getchar()) != EOF)    {        a[n] = &c[n];        c[n++] = ch;    }    c[n] = 0;    qsort(a, n, sizeof(char *), pstrcmp);    for (i = 0; i < n-1; i++)    {        temlen = comlen(a[i], a[i+1]);        if (temlen > maxlen)        {            maxlen = temlen;            maxi = i;        }    }    printf("%.*s\n", maxlen, a[maxi]);    return 0;}

正确实现二分查找算法

二分查找可以解决排序数组的查找问题：只要数组中包含T（即要查找的值），那么通过不断缩小包含T的范围，最终就可以找到它。一开始，范围覆盖整个数组。将数组的中间项与T进行比较，可以排除一半元素，范围缩小一半。就这样反复比较，反复缩小范围，最终就会在数组中找到T，或者确定原以为T所在的范围实际为空。对于包含N个元素的表，整个查找过程大约要经过log(2)N次比较。

//copyright@2011 July//随时欢迎读者找bug，email：zhoulei0907@yahoo.cn。//首先要把握下面几个要点：//right=n-1 => while(left <= right) => right=middle-1;//right=n   => while(left <  right) => right=middle;//middle的计算不能写在while循环外，否则无法得到更新。int binary_search(int array[],int n,int value){int left = 0;int right = n-1;//如果这里是int right = n 的话，那么下面有两处地方需要修改，以保证一一对应：//1、下面循环的条件则是while(left < right)//2、循环内当array[middle]>value 的时候，right = midwhile (left <= right)  //循环条件，适时而变{int middle = left + ((right-left)>>1);//防止溢出，移位也更高效。每次循环都需要更新。if (array[middle] > value){right = middle-1;   //right赋值，适时而变} else if(array[middle] < value){left = middle+1;}elsereturn middle;  //可能会有读者认为刚开始时就要判断相等，但毕竟数组中不相等的情况更多//如果每次循环都判断一下是否相等，将耗费时间}return -1;}