LeetCode Expression Add Operators

Description:

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 -> []

Solution:

一开始想的是区间，因为没法处理乘法，但是后来发现可以用DFS+记录前序数字……这个方法比较优雅

<span style="font-size:18px;">import java.util.*;

public class Solution {
List<String> list = new LinkedList<String>();
int n;
int target;

public List<String> addOperators(String num, int target) {
n = num.length();
this.target = target;
dfs(0, 0, num, "");
return list;
}

public void dfs(long pre, long sum, String str, String path) {
if (str.length() == 0) {
if (sum == target)
list.add(new String(path));
return;
}
for (int i = 1; i <= str.length(); i++) {
String currentStr = str.substring(0, i);
long currentNum = Long.parseLong(currentStr);
String nextStr = str.substring(i, str.length());
if (currentStr.charAt(0) == '0' && currentStr.length() > 1)
return;
if (path.length() > 0) {
dfs(currentNum, sum + currentNum, nextStr, path + "+"
+ currentStr);
dfs(-currentNum, sum - currentNum, nextStr, path + "-"
+ currentStr);
dfs(currentNum * pre, sum - pre + pre * currentNum, nextStr,
path + "*" + currentStr);
} else {
dfs(currentNum, currentNum, nextStr, currentStr);
}
}
}
}</span>