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: []

分析

解法是利用遞歸回溯來遍歷所有的可能，但是要註意一些邊界情形。

public class Solution {
    public List<String> addOperators(String num, int
 
 target) {
        List<String> rst = new ArrayList<String>();
        if(num == null || num.length() == 0) return rst;
        helper(rst, "", num, target, 0, 0, 0);
        return rst;
    }
　　// eval記錄當前計算結果，multed計算上次計算變化的部分，在選擇乘法時會用到這個
    public void helper(List<String> rst, String path, String num, int
 
 target, int pos, long eval, long multed){
        if(pos == num.length()){
            if(target == eval)
                rst.add(path);
            return;
        }
        for(int i = pos; i < num.length(); i++){
            if(i != pos && num.charAt(pos) == ‘0‘) break;　　// 拋棄以0開始的數字
            long cur = Long.parseLong(num.substring(pos, i + 1));
            if(pos == 0){
                helper(rst, path + cur, num, target, i + 1, cur, cur);  // 起始數字特殊處理
            }
            else{
                helper(rst, path + "+" + cur, num, target, i + 1, eval + cur , cur);　　// 對當前數字cur選擇加上之前的部分
                helper(rst, path + "-" + cur, num, target, i + 1, eval -cur, -cur);　　// 選擇減
　　　　　　　　// 選擇乘法要特殊處理，減去上次變化的部分，將這個變化的部分乘以當前的數字再加上去
                helper(rst, path + "*" + cur, num, target, i + 1, eval - multed + multed * cur, multed * cur );
            }
        }
    }
}

LeetCode282. Expression Add Operators