Letter Combinations of a Phone Number - LeetCode

子集问题，从多重循环到回溯

用一个path记录

回溯三问：

dfs(i)->dfs(i + 1)

这题要注意idx是我们遍历的数字的位数，backtracking的时候要到下一层就是下一个数字，每个数字都是不同得集合，这题是求不同集合得组合.

Time: O（n*4^n)

Space: O(n)

class Solution {
    String[] keypad = new String[]{"", "", "abc", "def", "ghi","jkl", "mno", "pqrs", "tuv", "wxyz"};

    public List<String> letterCombinations(String digits) {
        List<String> res = new ArrayList<>();
        if (digits == null || digits.length() == 0) return res;

        StringBuilder sb = new StringBuilder();
        backtracking(res, sb, digits, 0);
        return res;
    }

    private void backtracking(List<String> res, StringBuilder sb, String digits, int idx) {
        if (idx == digits.length()) {
            res.add(sb.toString());
            return;
        }

        String key = keypad[digits.charAt(idx) - '0'];
        for (int i = 0; i < key.length(); i++) {
            sb.append(key.charAt(i));
            backtracking(res, sb, digits, idx + 1);
            sb.deleteCharAt(sb.length() - 1);
        }
    }
}

Restore IP Addresses - LeetCode

判断isValid的地方，要注意细节

Time: O(3^4)

Space: O(n)

class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> res = new ArrayList<>();
        StringBuilder sb = new StringBuilder(s);
        backtracking(res, sb, 0, 0);
        return res;
    }

    private void backtracking(List<String> res, StringBuilder sb, int points, int idx) {
        if (points == 3) {
            if (isValid(sb, idx, sb.length() - 1)) {
                res.add(sb.toString());
            }
            return;
        }

        for (int i = idx; i < sb.length(); i++) {
            if (isValid(sb, idx, i)) {
                sb.insert(i + 1, '.');
                points += 1;
                backtracking(res, sb, points, i + 2);
                sb.deleteCharAt(i + 1);
                points -= 1;
            }
        }
    }

    private boolean isValid(StringBuilder sb, int left, int right) {
        if (left > right) return false;
        if (sb.charAt(left) == '0' && left != right) return false;
        int num = 0;
        for (int i = left; i <= right; i++) {
            if (sb.charAt(i) < '0' || sb.charAt(i) > '9') return false;
            int digit = sb.charAt(i) - '0';
            num = num * 10 + digit;
            if (num > 255) return false;
        }
        return true;

    }
}

N-Queens - LeetCode

因为在单层搜索的过程中，每一层递归，只会选for循环（也就是同一行）里的一个元素，所以不用去重了。

Time: O(n!)

Space: O(n)

class Solution {
    public List<List<String>> solveNQueens(int n) {
        List<List<String>> res = new ArrayList<>();
        char[][] chess = new char[n][n];
        for (char[] c : chess) {
            Arrays.fill(c, '.');
        }
        backtracking(res, chess, n, 0);
        return res;
    }

    private void backtracking(List<List<String>> res, char[][] chess, int n, int row) {
        if (row == n) {
            res.add(construct(chess));
            return;
        }

        for (int i = 0; i < n; i++) {
            if (isValid(row, i, n, chess)) {
                chess[row][i] = 'Q';
                backtracking(res, chess, n, row + 1);
                chess[row][i] = '.';
            }
        }
    }

    private  boolean isValid(int row, int col, int n, char[][] chess) {
        for (int i = 0; i < row; i++) {
            if (chess[i][col] == 'Q') return false;
        }

        for (int i = row - 1, j = col - 1; i >= 0 && j >= 0; i--, j--) {
            if (chess[i][j] == 'Q') return false;
        }

        for (int i = row - 1, j = col + 1; i >= 0 && j <= n - 1; i--, j++ ) {
            if (chess[i][j] == 'Q') return false;
        }
        return true;
    }

    private List<String> construct(char[][] chess) {
        List<String> path = new ArrayList<>();
        for (int i = 0; i < chess.length; i++) {
            path.add(new String(chess[i]));
        }
        return path;
    }
}