（算法题）Day-20230316

组合

class Solution {
    List<List<Integer>> res=new ArrayList<>();
    LinkedList<Integer> path=new LinkedList<>();
    public List<List<Integer>> combine(int n, int k) {
        backTracking(n,k,1);
        return res;
    }

    public void backTracking(int n,int k,int startIndex){
        if(path.size()==k){
            res.add(new ArrayList<>(path));
            return;
        }
        for(int i=startIndex; i<=n; i++){
            path.add(i);
            backTracking(n,k,i+1);
            path.removeLast();
        }
    }
}

组合总和 III

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    LinkedList<Integer> path = new LinkedList<>();
    public List<List<Integer>> combinationSum3(int k, int n) {
        backTracking(n,k,0,1);
        return res;
    }

    public void backTracking(int n,int k,int sum,int startIndex){
        if((sum==n) && (path.size()==k)){
            res.add(new ArrayList<>(path));
            return;
        }
        for(int i=startIndex;i<=9;i++){
            path.add(i);
            sum+=i;
            backTracking(n,k,sum,i+1);
            sum-=i;
            path.removeLast();
        }
    }
}

电话号码的字母组合

class Solution {
   List<String> res = new ArrayList<>();
   StringBuilder sb = new StringBuilder();

   public List<String> letterCombinations(String digits) {
      if (digits == null || digits.length() == 0) return res;
      String[] numString = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
      backTracking(digits, numString, 0);
      return res;
   }

   public void backTracking(String digits, String[] numString, int idx) {
      if (idx == digits.length()) {
         res.add(sb.toString());
         return;
      }
      String s = numString[digits.charAt(idx) - '0'];
      for (int i = 0; i < s.length(); i++) {
         sb.append(s.charAt(i));
         backTracking(digits, numString, idx + 1);
         sb.deleteCharAt(sb.length() - 1);
      }
   }
}

组合总和

public class Solution {

    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        ArrayList<Integer> path = new ArrayList<>();
        Arrays.sort(candidates);
        backTracking(res,path,candidates,target,0,0);
        return res;
    }

    public void backTracking(List<List<Integer>> res,ArrayList<Integer> path,int[] candidates,int target,int sum,int startIdx){
        if(target==sum){
            res.add(new ArrayList<>(path));
            return;
        }
        for(int i=startIdx;i<candidates.length;i++){
            if(sum+candidates[i]>target) break;
            path.add(candidates[i]);
            backTracking(res,path,candidates,target,sum+candidates[i],i);
            path.remove(path.size()-1);
        }
    }
}

组合总和II

注意去重逻辑！！！

class Solution {
  LinkedList<Integer> path = new LinkedList<>();
  List<List<Integer>> ans = new ArrayList<>();
  boolean[] used;
  int sum = 0;

  public List<List<Integer>> combinationSum2(int[] candidates, int target) {
    used = new boolean[candidates.length];
    // 加标志数组，用来辅助判断同层节点是否已经遍历
    Arrays.fill(used, false);
    // 为了将重复的数字都放到一起，所以先进行排序
    Arrays.sort(candidates);
    backTracking(candidates, target, 0);
    return ans;
  }

  private void backTracking(int[] candidates, int target, int startIndex) {
    if (sum == target) {
      ans.add(new ArrayList(path));
    }
    for (int i = startIndex; i < candidates.length; i++) {
      if (sum + candidates[i] > target) {
        break;
      }
      // 出现重复节点，同层的第一个节点已经被访问过，所以直接跳过
      if (i > 0 && candidates[i] == candidates[i - 1] && !used[i - 1]) {
        continue;
      }
      used[i] = true;
      sum += candidates[i];
      path.add(candidates[i]);
      // 每个节点仅能选择一次，所以从下一位开始
      backTracking(candidates, target, i + 1);
      used[i] = false;
      sum -= candidates[i];
      path.removeLast();
    }
  }
}