字符串的全排列

public class Main {
    
    static void permutation(char[] chars) {
        permutation(chars, 0, chars.length - 1);
    }

    /** 数组中从索引begin到索引end之间的子数组参与到全排列 */
    static void permutation(char[] chars, int begin, int end) {
        if (begin == end) { //只剩最后一个字符时为出口
            for (int i = 0; i < chars.length; ++i) {
                System.out.print(chars[i]);
            }
            System.out.println();
        } else {
            for (int i = begin; i <= end; i++) { //每个字符依次固定到数组或子数组的第一个
                if (canSwap(chars, begin, i)) { //判断是否重复了
                    swap(chars, begin, i); //
                    permutation(chars, begin + 1, end); //递归求子数组的全排列
                    swap(chars, begin, i); //还原
                }
            }
        }
    }
    
    //如果时重复字符串，就放弃交换，比如abb，第2次交换就没有必要了
    static boolean canSwap(char[] chars, int begin, int end) {
        for (int i = begin; i < end; ++i) {
            if (chars[i] == chars[end]) {
                return false;
            }
        }
        return true;
    }
    
    static void swap(char[] chars, int from, int to) {
        char temp = chars[from];
        chars[from] = chars[to];
        chars[to] = temp;
    }
    
    public static void main(String[] args) {
        final char[] chars = new char[] {‘a‘, ‘b‘, ‘c‘};
        permutation(chars);
    }
}