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 -> []
一刷
題解：類似于backtracking, 不過(guò)由于+/-和*的優(yōu)先級(jí)不同，于是引入一個(gè)mult的變量堡掏， 可以在遇到乘法的時(shí)候泉唁，用eval-mult得到乘法的左邊的數(shù)字亭畜。eval表示當(dāng)前的value
class Solution {
public List<String> addOperators(String num, int target) {
List<String> res = new ArrayList<>();
if(num == null || num.length() == 0) return res;
perm(res, num, "", 0, 0, 0, target);
return res;
}
private void perm(List<String> res, String num, String path, int pos, long eval, long mult, int target){
if(pos == num.length()){
if(eval == target) res.add(path);
return;
}
for(int i=pos; i<num.length(); i++){
long cur = Long.parseLong(num.substring(pos, i+1));//include i
if(i!=pos && num.charAt(pos) == '0') break;
if(pos == 0){
perm(res, num, path+cur, i+1, cur, cur, target);
}else{
perm(res, num, path + "+" + cur, i+1, eval+cur, cur, target);
perm(res, num, path + "-" + cur, i+1, eval-cur, -cur, target);
perm(res, num, path + "*" + cur, i+1, eval-mult + mult*cur, mult*cur, target);
}
}
}
}
