基數(shù)排序是在某種情況下比快速排序還快的排序.當(dāng)然了,計數(shù)排序(Counting Sort)也有可能比快速排序快. 計數(shù)排序非常容易理解,時間復(fù)雜度是O(MAX(a[i])), 如果數(shù)據(jù)范圍很小的話,計數(shù)排序有巨大優(yōu)勢. 而基數(shù)排序,則更進(jìn)一步,對每一位進(jìn)行計數(shù)排序. 這樣時間復(fù)雜度降為O(N*log(MAX(a[i])) 以下代碼實(shí)現(xiàn)了從小到大cntSort()和從大到小cntSort2().實(shí)際上也可以倒置得到從大到小,依然是O(N),代碼比較迷的地方就是output數(shù)組,記住循環(huán)順序,這個比較巧妙,具體參見 http://www.geeksforgeeks.org/radix-sort/
#include <bits/stdc++.h>using namespace std;int idx(int x, int exp){ return (x / exp) % 10;}void cntSort(int *a, int n, int exp){ int cnt[10] = {0}; int output[n]; for (int i = 0; i < n; i++) cnt[idx(a[i], exp)]++; for (int i = 1; i < 10; i++) cnt[i] += cnt[i - 1]; for (int i = n - 1; i >= 0; i--) { output[cnt[idx(a[i], exp)] - 1] = a[i]; cnt[idx(a[i], exp)]--; } for (int i = 0; i < n; i++) a[i] = output[i];}void cntSort2(int *a, int n, int exp){ int cnt[10] = {0}; int output[n]; for (int i = 0; i < n; i++) cnt[idx(a[i], exp)]++; for (int i = 8; i >= 0; i--) cnt[i] += cnt[i + 1]; for (int i = 0; i < n; i++) { output[cnt[idx(a[i], exp)] - 1] = a[i]; cnt[idx(a[i], exp)]--; } for (int i = 0; i < n; i++) a[i] = output[i];}int main(){ //freopen("in", "r", stdin); int n; scanf("%d", &n); int a[n]; int mx = 0; for (int i = 0; i < n; i++) { scanf("%d", &a[i]); mx = max(a[i], mx); } for (int exp = 1; mx / exp > 0; exp *= 10) cntSort(a, n, exp); for (int i = 0; i < n; i++) cout << a[i] << ' '; cout << endl; for (int exp = 1; mx / exp > 0; exp *= 10) cntSort2(a, n, exp); for (int i = 0; i < n; i++) cout << a[i] << ' ';}新聞熱點(diǎn)
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