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282. 给表达式添加运算符

English Version

题目描述

给定一个仅包含数字 0-9 的字符串和一个目标值，在数字之间添加 二元 运算符（不是一元）+、- 或 * ，返回所有能够得到目标值的表达式。

 

示例 1:

输入: num = "123", target = 6
输出: ["1+2+3", "1*2*3"] 

示例 2:

输入: num = "232", target = 8
输出: ["2*3+2", "2+3*2"]

示例 3:

输入: num = "105", target = 5
输出: ["1*0+5","10-5"]

示例 4:

输入: num = "00", target = 0
输出: ["0+0", "0-0", "0*0"]

示例 5:

输入: num = "3456237490", target = 9191
输出: []

 

提示：

• 0 <= num.length <= 10
• num 仅含数字

解法

Python3

class Solution:
    def addOperators(self, num: str, target: int) -> List[str]:
        ans = []

        def dfs(u, prev, curr, path):
            if u == len(num):
                if curr == target:
                    ans.append(path)
                return
            for i in range(u, len(num)):
                if i != u and num[u] == '0':
                    break
                next = int(num[u: i+1])
                if u == 0:
                    dfs(i + 1, next, next, path + str(next))
                else:
                    dfs(i + 1,  next, curr + next, path + "+" + str(next))
                    dfs(i + 1, -next, curr - next, path + "-" + str(next))
                    dfs(i + 1, prev * next, curr - prev +
                        prev * next, path + "*" + str(next))

        dfs(0, 0, 0, "")
        return ans

Java

class Solution {
    private List<String> ans;
    private String num;
    private int target;

    public List<String> addOperators(String num, int target) {
        ans = new ArrayList<>();
        this.num = num;
        this.target = target;
        dfs(0, 0, 0, "");
        return ans;
    }

    private void dfs(int u, long prev, long curr, String path) {
        if (u == num.length()) {
            if (curr == target) ans.add(path);
            return ;
        }
        for (int i = u; i < num.length(); i++) {
            if (i != u && num.charAt(u) == '0') {
                break;
            }
            long next = Long.parseLong(num.substring(u, i + 1));
            if (u == 0) {
                dfs(i + 1, next, next, path + next);
            } else {
                dfs(i + 1,  next, curr + next, path + "+" + next);
                dfs(i + 1, -next, curr - next, path + "-" + next);
                dfs(i + 1, prev * next, curr - prev + prev * next, path + "*" + next);
            }
        }
    }
}

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