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: []

題目

給定一個數字串S和一個值target，允許你在S中間新增加減乘符號，使得表示式結果為target，求所有添法。

思路

dfs + pruning（適當剪枝）

程式碼

 1   class Solution {
 2     public List<String> addOperators(String num, int target) {
 3         List<String> res = new ArrayList<>(); 
 4         dfs(num, 0, 0, 0, "", res, target);
 5         return res;
 6     }
 7      
 8     private
 
 void dfs(String num, int index, long sum, long last, String s, List<String> res, int target) {
 9             
10         if(index == num.length()) {
11             if(sum == target) {
12                 res.add(s);
13             }
14         }
15         
16         for(int i = index + 1; i <= num.length(); i++) {           
17             String temp = num.substring(index, i);
18             if(temp.length() > 1 && temp.charAt(0) == '0') {
19                 continue;
20             }
21             
22             long n = Long.valueOf(temp);
23             
24             if(index == 0) {
25                 dfs(num, i, sum + n, n, s + n, res, target);
26                 continue;
27             } 
28             dfs(num, i, sum + n, n,  s + "+" + n,  res, target);
29             dfs(num, i, sum - n, -n,  s + "-" + n, res, target);
30             dfs(num, i, (sum-last) + last * n, last * n, s  + "*" + n, res, target);
31         }
32     }
33 }