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;}
}