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java的Deque

Leetcode 39. 组合总和

题目：39. 组合总和
解析：代码随想录解析

解题思路

还是回溯三部曲

代码

class Solution {List<List<Integer>> res = new ArrayList<>();List<Integer> paths = new ArrayList<>();int curSum = 0;private void backtracking(int[] candidates, int target, int startIndex) {if (curSum >= target) {if (curSum == target)res.add(new ArrayList<>(paths));return;}for (int i = startIndex; i < candidates.length; i++) {paths.add(candidates[i]);curSum += candidates[i];backtracking(candidates, target, i);paths.remove(paths.size() -1);curSum -= candidates[i];}}public List<List<Integer>> combinationSum(int[] candidates, int target) {backtracking(candidates, target, 0);return res;}
}

总结

越来越熟练了

Leetcode 40. 组合总和 II

题目：40. 组合总和 II
解析：代码随想录解析

解题思路

递归加一定的减枝，如果第一个元素一样的话，就跳过。

代码

class Solution {List<List<Integer>> res = new ArrayList<>();List<Integer> paths = new ArrayList<>();int curSum = 0;private void backtracking(int[] candidates, int target, int startIndex) {if (curSum >= target) {if (curSum == target)res.add(new ArrayList<>(paths));return;}for (int i = startIndex; i < candidates.length; i++) {if (i > startIndex && candidates[i] == candidates[i-1])continue;paths.add(candidates[i]);curSum += candidates[i];backtracking(candidates, target, i + 1);paths.remove(paths.size() - 1);curSum -= candidates[i];}}public List<List<Integer>> combinationSum2(int[] candidates, int target) {Arrays.sort(candidates);backtracking(candidates, target, 0);return res;}
}

总结

暂无

Leetcode 131. 分割回文串

题目：131. 分割回文串
解析：代码随想录解析

解题思路

一个函数判断是否是回文
一个回溯函数

代码

class Solution {List<List<String>> res = new ArrayList<>();List<String> paths = new ArrayList<>();private boolean isPalindrome(String s, int begin, int end) {for (int i = begin, j = end; i < j; i++, j--) {if (s.charAt(i) != s.charAt(j))return false;}return true;}private void backtracking(String s, int startIndex){if (startIndex == s.length()) {res.add(new ArrayList<>(paths));return;}for (int i = startIndex; i < s.length(); i++) {if (isPalindrome(s, startIndex, i)) {paths.add(s.substring(startIndex, i+1));backtracking(s, i+1);paths.remove(paths.size()-1);}}}public List<List<String>> partition(String s) {backtracking(s, 0);return res;}
}

总结

暂无