[LeetCode] 282. Expression Add Operators

Given a string num that contains only digits and an integer target, return all possibilities to add the binary operators '+', '-', or '*' between the digits of num so that the resultant expression evaluates 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: []

Constraints:

• 1 <= num.length <= 10
• num consists of only digits.
• -231 <= target <= 231 - 1

给表达式添加运算符。

题意是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。

思路是DFS backtracking，一点点去尝试。helper函数里包括了如下几个参数

res - 最终的结果

path - 当前被构建的运算表达式

curNum - 当前被构建的数字

preNum - 之前一个被构建的数字

加减法比较好理解，这里乘法的部分稍微有点绕，我参考了这个帖子，帮助理解。乘法的结果 = curNum - preNum + preNum * curNum，乘法操作完毕之后，之前一个被构建的数字也就变成了preNum * curNum。

时间 - not sure

空间 - O(n)

Java实现

class Solution {
    public List<String> addOperators(String num, int target) {
        List<String> res = new ArrayList<>();
        // corner case
        if (num == null || num.length() == 0) {
            return res;
        }
        helper(res, "", num, target, 0, 0, 0);
        return res;
    }

    private void helper(List<String> res, String path, String num, int target, int pos, long curRes, long preNum) {
        // corner case
        if (pos == num.length()) {
            if (target == curRes) {
                res.add(path);
                return;
            }
        }
        for (int i = pos; i < num.length(); i++) {
            if (i != pos && num.charAt(pos) == '0') {
                return;
            }
            long curNum = Long.valueOf(num.substring(pos, i + 1));
            if (pos == 0) {
                helper(res, path + curNum, num, target, i + 1, curNum, curNum);
            } else {
                helper(res, path + "+" + curNum, num, target, i + 1, curRes + curNum, curNum);
                helper(res, path + "-" + curNum, num, target, i + 1, curRes - curNum, -curNum);
                helper(res, path + "*" + curNum, num, target, i + 1, curRes - preNum + preNum * curNum,
                        preNum * curNum);
            }
        }
    }
}

LeetCode 题目总结