[leetcode]282. Expression Add Operators 表达式添加运算符

Given a string that contains only digits 0-9 and a target value, return all possibilities to add binary operators (not unary) +, -, or * between the digits so they evaluate to the target value.

Example 1:

Input: num = "123", target = 6
Output: ["1+2+3", "1*2*3"] 

Example 2:

Input: num = "232", target = 8
Output: ["2*3+2", "2+3*2"]

Example 3:

Input: num = "105", target = 5
Output: ["1*0+5","10-5"]

Example 4:

Input: num = "00", target = 0
Output: ["0+0", "0-0", "0*0"]

Example 5:

Input: num = "3456237490", target = 9191
Output: []

题目

给定一个数字串S和一个值target，允许你在S中间添加加减乘符号，使得表达式结果为target，求所有添法。

思路

dfs + pruning（适当剪枝）

代码

  class Solution {
    public List<String> addOperators(String num, int target) {
        List<String> res = new ArrayList<>(); 
        dfs(num, 0, 0, 0, "", res, target);
        return res;
    }
     
    private void dfs(String num, int index, long sum, long last, String s, List<String> res, int target) {
            
        if(index == num.length()) {
            if(sum == target) {
                res.add(s);
            }
        }
        
        for(int i = index + 1; i <= num.length(); i++) {           
            String temp = num.substring(index, i);
            if(temp.length() > 1 && temp.charAt(0) == '0') {
                continue;
            }
            
            long n = Long.valueOf(temp);
            
            if(index == 0) {
                dfs(num, i, sum + n, n, s + n, res, target);
                continue;
            } 
            dfs(num, i, sum + n, n,  s + "+" + n,  res, target);
            dfs(num, i, sum - n, -n,  s + "-" + n, res, target);
            dfs(num, i, (sum-last) + last * n, last * n, s  + "*" + n, res, target);
        }
    }
}