Expression Add Operators(leetcode)

LeetCode题解：ExpressionAddOperators

最新推荐文章于 2020-03-09 23:19:16 发布

Valerie7012

最新推荐文章于 2020-03-09 23:19:16 发布

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分类专栏： leetcode 算法

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本文介绍了一个LeetCode上的编程挑战题“ExpressionAddOperators”，该题要求在给定的数字串中添加算术运算符，使得最终表达式的计算结果等于目标值。文中提供了一种递归回溯的方法来解决这个问题，并通过实例演示了如何实现。

Expression Add Operators

Expression Add Operators
- 题目
- 解决

题目

leetcode题目
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 -> []

解决

class Solution {
public:
    void addOperator(vector<string> & result, string nums, string r, long long last, long long current, int target) {
        if (nums.length() == 0) {
            if (current == target) {
                result.push_back(r);
            }
            return;
        }
        for (int i = 1; i <= nums.length(); i++) {
            string num = nums.substr(0, i);
            if (num.length() >= 2 && num[0] == '0') {
                return;
            }
            string next = nums.substr(i);
            if (r.length() > 0) {
                addOperator(result, next, r + "+" + num, stoll(num), current + stoll(num), target);
                addOperator(result, next, r + "-" + num, -stoll(num), current - stoll(num), target);
                addOperator(result, next, r + "*" + num, last * stoll(num), (current - last) + (last * stoll(num)), target);
            } else {
                addOperator(result, next, num, stoll(num), stoll(num), target);
            }
        }
    }

    vector<string> addOperators(string num, int target) {
        vector<string> result;
        addOperator(result, num, "", 0, 0, target);
        return result;
    }
};