10.字符串的不同排列

描述

给出一个字符串，找到它的所有排列，注意同一个字符串不要打印两次

样例

给出 "abb"，返回 ["abb", "bab", "bba"]。
给出 "aabb"，返回 ["aabb", "abab", "baba", "bbaa", "abba", "baab"]

代码

1. DFS

public class Solution {
    /*
     * @param str: A string
     * @return: all permutations
     */
    public List stringPermutation2(String str) {
        // write your code here
        List result = new ArrayList<>();
        
        if (str == null || str.length() == 0) {
            result.add("");
            return result;
        }
        
        //Sort Array beforehand for ordering purpose
        str = sortString(str);
        
        Set set = new HashSet<>(); 
        boolean[] isUsed = new boolean[str.length()]; 
        String testStr = "";
        
        // DFS
        DFS(str, testStr, isUsed,  set, result);
        return result;
    }
    
    private void DFS(String str, 
                     String testStr,
                     boolean[] isUsed,
                     Set set,
                     List result) {
        if (testStr.length() == str.length()) {
            result.add(new String(testStr)); 
            return;
        }
        
        for (int i = 0; i < str.length(); i++) {
            if (isUsed[i] == true || i != 0 && isUsed[i - 1] == false 
                && str.charAt(i) == str.charAt(i - 1)) {
                continue;
            }
            
            if (set.contains(testStr + str.charAt(i))) {
                continue;
            }
            
            // 字符串拼接的方法
            testStr = testStr + str.charAt(i); 
            set.add(testStr);
            isUsed[i] = true;
           
            DFS(str, testStr, isUsed,set, result);
            
            testStr = testStr.substring(0, testStr.length() - 1); 
            set.remove(testStr);
            isUsed[i] = false;
        }
    }
    
    private String sortString(String str) {
        char[] tempArray = str.toCharArray(); 
        Arrays.sort(tempArray);
        return new String(tempArray);
    }
}

2. 思路同下一个排列的题，从编号为0的排列不断寻找下一个排列，直到找到最后一个排列为止，把所有排列加入result

public class Solution {
    /**
     * @param str a string
     * @return all permutations
     */
    public List stringPermutation2(String str) {
        List result = new ArrayList();
        char[] s = str.toCharArray();
        
        Arrays.sort(s);
        // String.valueOf() 将括号中的值转换成字符串
        result.add(String.valueOf(s));
        
        while ((s = nextPermutation(s)) != null) {
            result.add(String.valueOf(s));
        }
        return result;
    }

    // 得到下一种组合
    public char[] nextPermutation(char[] nums) {
        int index = -1;
        for (int i = nums.length - 1; i > 0; i--) {
            if (nums[i] > nums[i - 1]){
                index = i - 1;
                break;
            }
        }
        // 表示当前是最后一个排列
        if (index == -1) {
            return null;
        }
        for (int i = nums.length - 1; i > index; i--) {
            if (nums[i] > nums[index]) {
                char temp = nums[i];
                nums[i] = nums[index];
                nums[index] = temp;
                break;
            }
        }
        reverse(nums, index + 1, nums.length - 1);
        return nums;
    }

    public void reverse(char[] num, int start, int end) {
        for (int i = start, j = end; i < j; i++, j--) {
            char temp = num[i];
            num[i] = num[j];
            num[j] = temp;
        }
    }
}