代码随想录算法训练营29期|day 25 任务以及具体安排

•  216.组合总和III
• 

  class Solution {
  	List<List<Integer>> result = new ArrayList<>();
  	LinkedList<Integer> path = new LinkedList<>();
  
  	public List<List<Integer>> combinationSum3(int k, int n) {
  		backTracking(n, k, 1, 0);
  		return result;
  	}
  
  	private void backTracking(int targetSum, int k, int startIndex, int sum) {
  		// 减枝
  		if (sum > targetSum) {
  			return;
  		}
  
  		if (path.size() == k) {
  			if (sum == targetSum) result.add(new ArrayList<>(path));
  			return;
  		}
  
  		// 减枝 9 - (k - path.size()) + 1
  		for (int i = startIndex; i <= 9 - (k - path.size()) + 1; i++) {
  			path.add(i);
  			sum += i;
  			backTracking(targetSum, k, i + 1, sum);
  			//回溯
  			path.removeLast();
  			//回溯
  			sum -= i;
  		}
  	}
  }

  思路：回溯的基本模板，关键是剪枝的操作，if(sum > targetSum)进行剪枝，9-（k - path.size()）+1进行剪枝。

•  17.电话号码的字母组合
• 

  class Solution {
  
      //设置全局列表存储最后的结果
      List<String> list = new ArrayList<>();
  
      public List<String> letterCombinations(String digits) {
          if (digits == null || digits.length() == 0) {
              return list;
          }
          //初始对应所有的数字，为了直接对应2-9，新增了两个无效的字符串""
          String[] numString = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
          //迭代处理
          backTracking(digits, numString, 0);
          return list;
  
      }
  
      //每次迭代获取一个字符串，所以会设计大量的字符串拼接，所以这里选择更为高效的 StringBuild
      StringBuilder temp = new StringBuilder();
  
      //比如digits如果为"23",num 为0，则str表示2对应的 abc
      public void backTracking(String digits, String[] numString, int num) {
          //遍历全部一次记录一次得到的字符串
          if (num == digits.length()) {
              list.add(temp.toString());
              return;
          }
          //str 表示当前num对应的字符串
          String str = numString[digits.charAt(num) - '0'];
          for (int i = 0; i < str.length(); i++) {
              temp.append(str.charAt(i));
              //c
              backTracking(digits, numString, num + 1);
              //剔除末尾的继续尝试
              temp.deleteCharAt(temp.length() - 1);
          }
      }
  }

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

  思路：要把数字对应为字符串，将num作为遍历第几个数字的标识，递归num++，往深遍历，for循环是往宽遍历，遍历一个数字对应的所有字符