282. 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 -> []

一刷
题解：类似于backtracking, 不过由于+/-和*的优先级不同，于是引入一个mult的变量， 可以在遇到乘法的时候，用eval-mult得到乘法的左边的数字。eval表示当前的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);
            }
        }
    }
}