784. Letter Case Permutation

Description

Given a string S, we can transform every letter individually to be lowercase or uppercase to create another string. Return a list of all possible strings we could create.

Examples:
Input: S = "a1b2"
Output: ["a1b2", "a1B2", "A1b2", "A1B2"]

Input: S = "3z4"
Output: ["3z4", "3Z4"]

Input: S = "12345"
Output: ["12345"]

Note:

• S will be a string with length at most 12.
• S will consist only of letters or digits.

Solution

DFS, time O(2 ^ k), space ?

自底向上的写法

class Solution {
    public List letterCasePermutation(String S) {
        return dfsPermutation(S.toLowerCase(), 0);
    }
    
    public List dfsPermutation(String s, int start) {
        List res = new ArrayList<>();
        
        if (start == s.length()) {
            res.add("");
            return res;
        }
        
        char c = s.charAt(start);
        List sub = dfsPermutation(s, start + 1);
        for (String permutation : sub) {
            res.add(c + permutation);
        }
        
        if (Character.isLetter(c)) {
            c = Character.toUpperCase(c);
            for (String permutation : sub) {
                res.add(c + permutation);
            }
        }
        
        return res;
    }
}

DFS, time O(2 ^ k), space O(n)

自顶向下的写法

class Solution {
    public List letterCasePermutation(String s) {
        if (s == null) {
            return Collections.EMPTY_LIST;
        }
        
        s = s.toLowerCase();
        List res = new ArrayList<>();
        StringBuilder sb = new StringBuilder();
        dfs(s, 0, sb, res);
        return res;
    }
    
    public void dfs(String s, int i, StringBuilder sb, List res) {
        if (i == s.length()) {
            res.add(sb.toString());
            return;
        }
        
        sb.append(s.charAt(i));
        dfs(s, i + 1, sb, res);
        if (Character.isLetter(s.charAt(i))) {
            sb.setCharAt(i, Character.toUpperCase(s.charAt(i)));
            dfs(s, i + 1, sb, res);
        }
        sb.deleteCharAt(sb.length() - 1);
    }
}

Bit-manipulation

注意细节，容易出错。

class Solution {
    public List letterCasePermutation(String s) {
        s = s.toLowerCase();
        List res = new ArrayList<>();
        int letterCount = 0;
        char[] arr = s.toCharArray();
        
        for (char c : arr) {
            if (Character.isLetter(c)) {
                ++letterCount;
            }
        }
        
        for (int bits = 0; bits < (1 << letterCount); ++bits) {
            StringBuilder sb = new StringBuilder();
            int mask = 1 << (letterCount - 1);  // important to use a mask!
            
            for (char c : arr) {
                if (Character.isLetter(c)) {
                    if ((bits & mask) == 0) {
                        sb.append(c);
                    } else {
                        sb.append(Character.toUpperCase(c));
                    }
                    
                    mask >>= 1;     // change mask, not bits
                } else {
                    sb.append(c);
                }
            }
            
            res.add(sb.toString());
        }
        
        return res;
    }
}