LeetCode-Expression Add Operators

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.

Examples:

"123", 6 -> ["1+2+3", "1*2*3"] 
"232", 8 -> ["2*3+2", "2+3*2"]
"105", 5 -> ["1*0+5","10-5"]
"00", 0 -> ["0+0", "0-0", "0*0"]
"3456237490", 9191 -> []

Credits:
Special thanks to @davidtan1890 for adding this problem and creating all test cases.

Analysis:

Use recursion+backtracking. We need deal with

1. Numbers cannot start with 0

2. When do multiply, say previous value is x = a+b*c, in current layer, we get number d and do multiplying, the value should be x-b*c+b*c*d. So we need to carry the information of the last mulitplied number into next layer, i.e., b*c*d.

 1 public class Solution {
 2     public List<String> addOperators(String num, int target) {
 3         List<String> resList = new ArrayList<String>();
 4         StringBuilder builder = new StringBuilder();
 5         addOperRecur(resList, num, target, builder, 0, 0, 0);
 6         return resList;
 7     }
 8 
 9     public void addOperRecur(List<String> resList, String num, int target, StringBuilder builder, int pos, long value,
10             long lastMultipledNum) {
11         if (pos >= num.length() && value == target) {
12             resList.add(builder.toString());
13             return;
14         }
15 
16         int builderLen = builder.length();
17         for (int i = pos; i < num.length(); i++) {
18             // only allow "0", not "0XXX".
19             if (i > pos && num.charAt(pos) == '0')
20                 return;
21             long curNum = Long.parseLong(num.substring(pos, i + 1));
22             if (pos == 0) {
23                 builder.append(curNum);
24                 addOperRecur(resList, num, target, builder, i + 1, curNum, curNum);
25                 builder.setLength(builderLen);
26             } else {
27                 builder.append('+').append(curNum);
28                 addOperRecur(resList, num, target, builder, i + 1, value + curNum, curNum);
29                 builder.setLength(builderLen);
30 
31                 builder.append('-').append(curNum);
32                 addOperRecur(resList, num, target, builder, i + 1, value - curNum, -curNum);
33                 builder.setLength(builderLen);
34 
35                 builder.append('*').append(curNum);
36                 addOperRecur(resList, num, target, builder, i + 1, value - lastMultipledNum + lastMultipledNum * curNum,
37                         lastMultipledNum * curNum);
38                 builder.setLength(builderLen);
39             }
40         }
41     }
42 }
原文地址:https://www.cnblogs.com/lishiblog/p/5824410.html