代码随想录算法训练营29期|day 25 任务以及具体安排

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。

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