LeetCode刷题--Letter Case Permutat

题目

原题地址

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.

答题

row1: abc
row2: abc Abc
row3: abc aBc Abc ABc
row4: abc abC aBc aBC Abc AbC ABc ABC

有点像我们高中学的排列组合，我们能够知道一共有多少种组合，也知道每一个字母有两种情况，每一个数字有一种情况。

如上图，给出 abc， 我们已经知道有8种组合，根据上一行的组合，只改动一个位置上的字母即可的到本次的组合。以此类推，即可得到最终的所有组合。

另外此题是在考 BFS/DFS 算法，有兴趣的同学可以自行搜索学习。

public List<String> letterCasePermutation(String S) {
    if (S == null || S.length() == 0) {
        return new ArrayList<String>(Arrays.asList(""));
    }

    char[] array = S.toCharArray();
    int arraySize = array.length;

    int count = 0;
    for (char ch : array) {
        if (Character.isLetter(ch)) {
            count++;
        }
    }

    if (count == 0) {
        return new ArrayList<String>(Arrays.asList(S));
    }

    List<String> list = new ArrayList<String>(2 << (count - 1));
    list.add(S);
    for (int i = 0; i < arraySize; i++) {
        char ch = array[i];
        if (Character.isDigit(ch)) {
            continue;
        }
        int size = list.size();
        for (int j = 0; j < size; j++) {
            char[] temp = list.get(j).toCharArray();
            if (Character.isUpperCase(ch)) {
                temp[i] = Character.toLowerCase(ch);
            } else {
                temp[i] = Character.toUpperCase(ch);
            }
            list.add(String.valueOf(temp));
        }
    }
    return list;
}