leetcode46 全排列

使用回溯思想
for循环遍历每次选取的数
递归遍历下一次选取的数
选取完回溯 将暂时保存的path删除尾部 将used重置为0

class Solution {List<List<Integer>> res;List<Integer> path;public List<List<Integer>> permute(int[] nums) {res = new ArrayList<>();path = new ArrayList<>();backtracking(nums,new int[nums.length]);return res;}public void backtracking(int[] nums,int[] used){if(path.size() == nums.length){res.add(new ArrayList<>(path));return;}for(int i=0;i<nums.length;i++){if(used[i] ==1) continue;path.add(nums[i]);used[i]=1;backtracking(nums,used);path.removeLast();used[i]=0;}}
}

leetcode 78. 子集

考虑添加结果的时间点 不是在递归出口， 而是在每次更新path的时候，每次递归会使得横向遍历的结果缩小变成i+1，回溯阶段删除path的末尾

class Solution {List<List<Integer>> res;List<Integer> path ;public List<List<Integer>> subsets(int[] nums) {res = new ArrayList<>();path = new ArrayList<>();res.add(path);backtracking(nums,0);return res;}public void backtracking(int[] nums,int startIndex){if(startIndex >=nums.length){return;}for(int i = startIndex;i<nums.length;i++){path.add(nums[i]);res.add(new ArrayList(path));backtracking(nums,i+1);path.removeLast();}}}

leetcode 17. 电话号码的字母组合

横向for循环每个按钮里的值
递归下一层则是换一个按钮
每次都是从0开始，digit的长度为几就递归深度为几

class Solution {List<String> res;StringBuilder sb;public List<String> letterCombinations(String digits) {String[] strs = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};res = new ArrayList<>();if(digits.length()<=0) return res;sb  = new StringBuilder();backTracking(digits,0,strs);return res;}public void backTracking(String digits,int index,String[] strs){if(index >= digits.length()){res.add(sb.toString());return;}String temp = strs[digits.charAt(index)-'0'];for(int i =0;i<temp.length();i++){sb.append(temp.charAt(i));backTracking(digits,index+1,strs);sb.deleteCharAt(sb.length()-1);}}
}

leetcode 39. 组合总和

递归出口为结果等于target，如果大于target也退出递归 因为 输入数字串是从小到大排列
第一层递归要从当前值取起，但是每一层for循环会使i+1
回溯 删除结果末尾的值

class Solution {List<List<Integer>> res;List<Integer> path;public List<List<Integer>> combinationSum(int[] candidates, int target) {res = new ArrayList<>();path = new ArrayList<>();backTracking(candidates,target,0,0);return res;}public void backTracking(int[] candidates,int target,int total,int index){if(total>target){return;}if(total== target){res.add(new ArrayList<>(path));return;}for(int i =index;i<candidates.length;i++){path.add(candidates[i]);backTracking(candidates,target,total+candidates[i],i);path.removeLast();}}
}

leetcode 22. 括号生成

待取值数组为（），每一次向下递归不会改变，for循环01 0为（ 1 为），递归出口 l》n||r》n||r>l则直接退出不保存结果
lr && ln 则保存结果并退出

class Solution {List<String> res;StringBuilder sb = new StringBuilder();public List<String> generateParenthesis(int n) {res = new ArrayList<>();backTracking(0,0,n);return res;}public void backTracking(int l,int r,int n){if(r>n || l>n || r>l){return;}if(l==r && l==n){res.add(sb.toString());}for(int i =0;i<2;i++){if(i==0){sb.append("(");backTracking(l+1,r,n);} else if(i==1){sb.append(")");backTracking(l,r+1,n);}sb.deleteCharAt(sb.length()-1);}}
}