基數排序
算法(遞歸實現):
package com.zq.algorithm.sort;
/**
* Created by zhengshouzi on 2015/11/2.
*/
public class RadixSort {
public static void main(String[] args) {
int[] a = {135, 242, 192, 93, 345, 11, 24, 19};
redixSort(a, 3, 1);
System.out.println("最終結果:");
for (int i = 0; i < a.length; i++) {
System.out.print(" " + a[i]);
}
}
public static void redixSort(int[] a, int digit, int divide) {
//遞歸退出條件,Math.pow()是API裏面求冪的函數,這裏是10的digit次方
if (divide < Math.pow(10, digit)) {
//基數
int redix = 10;
//一個零時數組,零時存放原來的數組,一個記錄桶信息,
int[] temp = new int[a.length];
int[] count = new int[redix];
//這個循環就是生成每個桶需要記錄的信息
for (int j = 0; j < a.length; j++) {
//System.out.println(a[j]);
int tempKey = a[j] / divide % redix;
count[tempKey]++;
}
//這個循環是拷貝數組,當然API有自帶拷貝數組的API,爲了說明問題,就不用了。
for (int i = 0; i < a.length; i++) {
temp[i] = a[i];
}
```下面的註釋是,一開始我沒有使用計數排序,自己拍腦袋想的方法```
/*
//定義一個變量,用來控制 a 數組下標
int j = 0;
//循環找到通過每一個基數排序後的順序,
for (int k = 0; k < count.length; k++) {
//如果存在基數對應的數,
if (count[k] != 0) {
//遍歷temp數組,將和K值對應的那個數找出來,順序放入到 a 數組中,這樣就完成了 某一位上面的排序。
for (int l = 0; l < temp.length; l++) {
int tempKey = (temp[l] / divide) % redix;
if (tempKey == k) {
a[j++] = temp[l];
}
}
}
}*/
for (int i = 1; i < redix; i++) {
count [i] = count[i] + count[i-1];
}
//計數排序
for (int i = a.length-1; i >= 0; i--) {
int tempKey = (temp[i]/divide)%redix;
count[tempKey]--;
a[count[tempKey]] = temp[i];
}
/*
for (int i = 0; i < count.length; i++) {
System.out.print(" " + count[i]);
}
*/
System.out.println();
System.out.print("對" + divide + "位計數排序後:");
for (int i = 0; i < a.length; i++) {
System.out.print(" " + a[i]);
}
System.out.println();
//遞歸調用
redixSort(a, digit, divide * 10);
}
}
}