22. Generate Parentheses - Medium
阿新 • • 發佈:2018-12-07
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
For example, given n = 3, a solution set is:
[ "((()))", "(()())", "(())()", "()(())", "()()()" ]
M1: backtracking
只用一個stringbuilder,每一層遞迴完,刪掉最後一個char,回到上一層
time: O(2^n), space: O(n)
class Solution { public List<String> generateParenthesis(int n) { List<String> res = new ArrayList<>(); backtracking(n, n, new StringBuilder(), res); return res; } private void backtracking(int l, int r, StringBuilder sb, List<String> res) {if(l > r || l < 0 || r < 0) return; if(l == 0 && r == 0) res.add(sb.toString()); if(l > 0) { backtracking(l - 1, r, sb.append("("), res); sb.deleteCharAt(sb.length() - 1); } if(r > 0) { backtracking(l, r- 1, sb.append(")"), res); sb.deleteCharAt(sb.length() - 1); } } }
M2: dfs
符合條件的能被加入res中的字串,有n個 ( 和n個),當sb的長度為2n時,加入res。在recursion的時候,判斷string是否符合條件,可以用兩個引數l, r表示 ( 和 ) 剩餘的個數,如果遞迴過程中,字串中)的個數大於(的個數,即l > r,字串不符合條件;否則,當 l 和 r 都 = 0時,把字串加入res。
time: O(2^n), space: O(n)
class Solution { public List<String> generateParenthesis(int n) { List<String> res = new ArrayList<>(); backtracking(n, n, "", res); return res; } private void backtracking(int l, int r, String s, List<String> res) { if(l > r || l < 0 || r < 0) return; if(l == 0 && r == 0) res.add(s); if(l > 0) backtracking(l - 1, r, s + "(", res); if(r > 0) backtracking(l, r - 1, s + ")", res); } }