leetcode刷题之回溯法

Posted 2021-12-04 Panda_Java

掌握回溯法模板

• leetcode17 电话号码的字母组合
• leetcode78 子集

leetcode17 电话号码的字母组合

class Solution {

    public List<String> letterCombinations(String digits) {

        HashMap<Character,String> map = new HashMap<>();
        map.put('2',"abc");    
        map.put('3', "def");
        map.put('4', "ghi");
        map.put('5', "jkl");
        map.put('6', "mno");
        map.put('7', "pqrs");
        map.put('8', "tuv");
        map.put('9', "wxyz");

        List<String> res = new ArrayList<>();

        if(digits == null || digits.length() == 0)
            return res;

        StringBuilder cur = new StringBuilder();
        // 字符串拼接 StringBuilder(没有涉及多线程安全问题) StringBuffer 
        backtracking(digits,map,0,res,cur);

        return res;
    }

// 不需要返回结果集
    private void backtracking(String digits, HashMap<Character,String> map, int index, List<String> res, StringBuilder str){  //遍历字符串digits每个位置 start表示当前索引位置  每一个组合用StringBuilder来装
        if(index == digits.length())
            res.add(str.toString());
        else{
            String s = map.get(digits.charAt(index));
            int count =  s.length();
            for(int i = 0; i <count; i++){
                str.append(s.charAt(i));
                backtracking(digits,map,index+1,res,str);
                //str.remove(str.length()-1);
                str.deleteCharAt(index);
            }
        }
    }
}

leetcode78 子集

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
      //回溯法
      List<Integer> list = new LinkedList<>();
      List<List<Integer>> ans = new LinkedList<>();
      ans.add(list);
      for(int i = 1; i <= nums.length; i++){
          backTracking(nums,i,0,ans,new LinkedList<>());
      }
      return ans;
    }
    private void backTracking(int[] nums, int length, int index, List<List<Integer>> ans,  List<Integer> subset){
        if(subset.size() == length){
            List<Integer> temp = new LinkedList<>(subset);
            ans.add(temp);
            return;
        }
        for(int i = index; i < nums.length; i++){
            subset.add(nums[i]);
            backTracking(nums,length,i+1,ans,subset);
            subset.remove(subset.size()-1);
        }
    }
}