代码随想录第二十八天

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复习

77 组合

注意剪枝操作，以及for loop的判断条件

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

    private void getPath(int n, int k, int start){
        if (path.size() == k){
            result.add(new ArrayList<>(path));
            return;
        }

        for (int i = start; i <= n - (k - path.size()) + 1; i++){
            path.add(i);
            getPath(n, k, i + 1);
            path.remove(path.size() - 1);
        }
    }
}

17 电话号码的字母组合

注意for loop之前的值

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class Solution {
    String[] numString = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};

    List<String> result = new ArrayList<>();
    StringBuilder sb = new StringBuilder();
    public List<String> letterCombinations(String digits) {
        if (digits == null || digits.length() == 0) return result;
        backTracking(digits, 0);
        return result;
    }

    private void backTracking(String digits, int start){
        if (sb.length() == digits.length()){
            result.add(sb.toString());
            return;
        }

        String str = numString[digits.charAt(start) - '0'];
        for (int i = 0; i < str.length(); i++){
            sb.append(str.charAt(i));
            backTracking(digits, start + 1);
            sb.deleteCharAt(sb.length() - 1);
        }
    }
}

39 组合总和

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class Solution {
    List<List<Integer>> result = new ArrayList<>();
    List<Integer> path = new ArrayList<>();

    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        if (candidates.length == 0) return result;
        backTracking(candidates, target, 0);
        return result;    
    }

    private void backTracking(int[] candidates, int target, int start){
        if (target < 0){
            return;
        }
        if (target == 0){
            result.add(new ArrayList<>(path));
            return;
        }

        for (int i = start; i < candidates.length; i++){
            target -= candidates[i];
            path.add(candidates[i]);
            backTracking(candidates, target, i);
            path.remove(path.size() - 1);
            target += candidates[i];
        }
    }
}

40 组合总和II

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class Solution {
    List<List<Integer>> result = new ArrayList<>();
    List<Integer> path = new ArrayList<>();
    boolean[] used;

    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        used = new boolean[candidates.length];
        Arrays.fill(used, false);
        Arrays.sort(candidates);
        backTracking(candidates, target, 0);  
        return result;  
    }

    private void backTracking(int[] candidates, int target, int start){
        if (target == 0){
            result.add(new ArrayList<>(path));
            return;
        }

        for (int i = start; i < candidates.length; i++){
            if (target < 0) break;
            if (i > 0 && candidates[i] == candidates[i - 1] && !used[i-1]) continue;

            target -= candidates[i];
            path.add(candidates[i]);
            used[i] = true;
            backTracking(candidates, target, i + 1);
            path.remove(path.size() - 1);
            target += candidates[i];
            used[i] = false;
        }
    }
}

216 组合总和III

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

    private void backTracking(int k, int n, int start){
        if (n < 0) return;
        if (n == 0 && path.size() == k){
            result.add(new ArrayList<>(path));
            return;
        }

        for (int i = start; i <= 9 - (k - path.size()) + 1; i++){
            n -= i;
            path.add(i);
            backTracking(k, n, i + 1);
            path.remove(path.size() - 1);
            n += i;
        }
    }
}

131 分割回文串

分割过程的for loop
先是切割出一个字符判断，如果是，就切割第二个，是切割第三个，完成后
第二个循环开始，直接切割两个，是移动，不是就继续第三个循环

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class Solution {
    List<List<String>> result = new ArrayList<>();
    List<String> path = new ArrayList<>();
    public List<List<String>> partition(String s) {
        backTracking(s, 0);
        return result;
    }

    private void backTracking(String s, int start){
        if (start == s.length()){
            result.add(new ArrayList<>(path));
            return;
        }

        for (int i = start; i < s.length(); i++){
            if (isPalindrome(s, start, i)){
                path.add(s.substring(start, i+1));
                backTracking(s, i + 1);
                path.remove(path.size() - 1);
            }
        }
    }
    private boolean isPalindrome(String s, int left, int right){
        while (left <= right){
            if (s.charAt(left) != s.charAt(right)) return false;
            left++;
            right--;
        }

        return true;
    }
}

93 复原IP 地址

分割问题和分割回文串类似
几个重点，

  1. 代码随想录利用了一个pointnum记录逗号数量，判断
    我没有用pointnum
    我用Integer.parseInt()转化了string
  2. 在实际问题中，判断数字是否valid也很重要
    1. 数字起始不能有0
    2. 数字要在0 到 255中间
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      class Solution {
          List<String> result = new ArrayList<>();
          List<Integer> path = new ArrayList<>();
          public List<String> restoreIpAddresses(String s) {
              backTracking(s, 0);
              return result;
          }

          private void backTracking(String s, int start){
              if (path.size() == 4){
                  StringBuilder sb = new StringBuilder();
                  for (int i = 0; i < path.size() - 1; i++){
                      sb.append(path.get(i)).append(".");
                  }
                  sb.append(path.get(path.size() - 1));
                  if (sb.length() == s.length() + 3){
                      result.add(sb.toString());
                  }
                  return;
              }

              for (int i = start; i < s.length(); i++){
                  if (isValid(s, start, i)){
                      path.add(Integer.parseInt(s.substring(start, i + 1)));
                      backTracking(s, i + 1);
                      path.remove(path.size() - 1);
                  } else {
                      break;
                  }
                  
              }
          }

          private boolean isValid(String s, int start, int end){
              if (start > end){
                  return false;
              }
              if (s.charAt(start) == '0' && end != start) return false;

              int num = 0;
              for (int i = start; i <= end; i++){
                  if (s.charAt(i) > '9' || s.charAt(i) < '0'){
                      return false;
                  }

                  num = num * 10 + (s.charAt(i) - '0');
                  if (num > 255){
                      return false;
                  }
              }

              return true;
          }
      }

78 子集

与组合和分割都不一样，
如果把子集问题抽象成树，就要把所有结点都记录下来

终止条件，就是start到了nums length的时候

result一直添加path

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class Solution {
    List<List<Integer>> result = new ArrayList<>();
    List<Integer> path = new ArrayList<>();
    public List<List<Integer>> subsets(int[] nums) {
        backTracking(nums, 0);
        return result;
    }

    private void backTracking(int[] nums, int start){
        result.add(new ArrayList<>(path));
        if (start >= nums.length) return;

        for (int i = start; i < nums.length; i++){
            path.add(nums[i]);
            backTracking(nums, i + 1);
            path.remove(path.size() - 1);
        }
    }
}

90 子集II

利用组合中的去重原理

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class Solution {
    List<List<Integer>> result = new ArrayList<>();
    List<Integer> path = new ArrayList<>();
    boolean[] used;
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        used = new boolean[nums.length];
        Arrays.sort(nums);
        Arrays.fill(used, false);
        backTracking(nums, 0);
        return result;
    }

    private void backTracking(int[] nums, int start){
        result.add(new ArrayList<>(path));
        if (start > nums.length){
            return;
        }

        for (int i = start; i < nums.length; i++){
            if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) continue;
            used[i] = true;
            path.add(nums[i]);
            backTracking(nums, i +1);
            used[i] = false;
            path.remove(path.size() - 1);
        }
    }
}

拓展 645 setmismatch

思路
利用loop循环查找，不等于同位置的结点，将该节点放到对应的位置上，进行排序，然后再利用for loop查找不等于该位置上的值输出result

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class Solution {
    public int[] findErrorNums(int[] nums) {
        int[] result = new int[2];
        int i = 0;
        while (i < nums.length){
            if (nums[nums[i] - 1] != nums[i]){
                swap(nums, nums[i] - 1, i);
            } else {
                i++;
            }
        }

        for (i = 0; i < nums.length; i++){
            if (nums[i] != i + 1){
                result[0] = nums[i];
                result[1] = i + 1;
            }
        }

        return result;
    }

    private void swap(int[] nums, int index1, int index2){
        nums[index1] ^= nums[index2];
        nums[index2] ^= nums[index1];
        nums[index1] ^= nums[index2];
    }
}