Given a string, find all permutations of it without duplicates.

Example

Given "abb", return ["abb", "bab", "bba"].

Given "aabb", return ["aabb", "abab", "baba", "bbaa", "abba", "baab"].

函數頭：public List<String> stringPermutation2(String str)

算法：DFS。和排列數字差不多，只是操作對象變了。private void dfs(char[] chars, boolean[] isUsed, StringBuilder sb, List<String> result) 。排序，遞歸函數。每次試著加一個字母，標記被用過，再遞歸。為了去重記得要求如果前後兩個char一樣，而且前面老大沒被用過，那你就不能被加。

細節：1.stringBuilder.deleteCharAT(idx) 可以去除某一位的char(切記是delete不是remove！更有操作小東西的感覺)。2.stringBuilder.length() 直接就是它當前包含的字符串的char數。3.stringBuilder.toString()相當於deepcopy，放心，你這時候照的相不會受之後sb變動而改變。4.初始化數組也要加new關鍵詞啊 int[] a = new int[5]。要是沒有new那還以為你在引用呢。

public class Solution {
    /*
     * @param str: A string
     * @return: all permutations
     
 
*/
    public List<String> stringPermutation2(String str) {
        // write your code here
        
        List<String> result = new ArrayList<>();
        if (str == null) {
            return result;
        }
        
        char[] chars = str.toCharArray();
        boolean[] isUsed = new
 
 boolean[chars.length];
        // 一定要排序哦
        Arrays.sort(chars);
        dfs(chars, isUsed, new StringBuilder(), result);
        
        return result;
    }
    
    private void dfs(char[] chars, boolean[] isUsed, StringBuilder sb, List<String> result) {
        
        if (sb.length() == chars.length) {
            result.add(sb.toString());
            return;
        }
        
        for (int i = 0; i < chars.length; i++) {
            if (isUsed[i]) {
                continue;
            }
            if (i > 0 && chars[i] == chars[i - 1] && !isUsed[i - 1]) {
                continue;
            }
            sb.append(chars[i]);
            isUsed[i] = true;
            dfs(chars, isUsed, sb, result);
            isUsed[i] = false;
            sb.deleteCharAt(sb.length() - 1);
        }
        
    }
}

lintcode10- String Permutation II- medium