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[LeetCode] Expression Add Operators 表达式增加操作符

 

Given a string that contains only digits 0-9 and a target value, return all possibilities to add binaryoperators (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: []

Credits:
Special thanks to @davidtan1890 for adding this problem and creating all test cases.

 

这道题给了我们一个只由数字组成的字符串,让我们再其中添加+,-或*号来形成一个表达式,该表达式的计算和为给定了target值,让我们找出所有符合要求的表达式来。看了题目中的例子1和2,很容易让人误以为是必须拆成个位数字,其实不是的,比如例子3中的 "105", 5能返回"10-5",说明连着的数字也可以。如果非要在过往的题中找一道相似的题,我觉得跟 Combination Sum II 很类似。不过这道题要更复杂麻烦一些。还是用递归来解题,我们需要两个变量diff和curNum,一个用来记录将要变化的值,另一个是当前运算后的值,而且它们都需要用 long 型的,因为字符串转为int型很容易溢出,所以我们用长整型。对于加和减,diff就是即将要加上的数和即将要减去的数的负值,而对于乘来说稍有些复杂,此时的diff应该是上一次的变化的diff乘以即将要乘上的数,有点不好理解,那我们来举个例子,比如 2+3*2,即将要运算到乘以2的时候,上次循环的 curNum = 5, diff = 3, 而如果我们要算这个乘2的时候,新的变化值diff应为 3*2=6,而我们要把之前+3操作的结果去掉,再加上新的diff,即 (5-3)+6=8,即为新表达式 2+3*2 的值,有点难理解,大家自己一步一步推算吧。

还有一点需要注意的是,如果输入为"000",0的话,容易出现以下的错误:

Wrong:["0+0+0","0+0-0","0+0*0","0-0+0","0-0-0","0-0*0","0*0+0","0*0-0","0*0*0","0+00","0-00","0*00","00+0","00-0","00*0","000"]

Correct:["0*0*0","0*0+0","0*0-0","0+0*0","0+0+0","0+0-0","0-0*0","0-0+0","0-0-0"]

我们可以看到错误的结果中有0开头的字符串出现,明显这不是数字,所以我们要去掉这些情况,过滤方法也很简单,我们只要判断长度大于1且首字符是‘0’的字符串,将其滤去即可,参见代码如下:

 

class Solution {
public:
    vector<string> addOperators(string num, int target) {
        vector<string> res;
        helper(num, target, 0, 0, "", res);
        return res;
    }
    void helper(string num, int target, long diff, long curNum, string out, vector<string>& res) {
        if (num.size() == 0 && curNum == target) {
            res.push_back(out); return;
        }
        for (int i = 1; i <= num.size(); ++i) {
            string cur = num.substr(0, i);
            if (cur.size() > 1 && cur[0] == '0') return;
            string next = num.substr(i);
            if (out.size() > 0) {
                helper(next, target, stoll(cur), curNum + stoll(cur), out + "+" + cur, res);
                helper(next, target, -stoll(cur), curNum - stoll(cur), out + "-" + cur, res);
                helper(next, target, diff * stoll(cur), (curNum - diff) + diff * stoll(cur), out + "*" + cur, res);
            } else {
                helper(next, target, stoll(cur), stoll(cur), cur, res);
            }
        }
    }
};

 

类似题目:

Evaluate Reverse Polish Notation

Basic Calculator II

Basic Calculator

Different Ways to Add Parentheses

Target Sum

 

参考资料:

https://leetcode.com/problems/expression-add-operators/

https://leetcode.com/problems/expression-add-operators/discuss/71971/Accepted-C%2B%2B-Solution

https://leetcode.com/problems/expression-add-operators/discuss/71895/Java-Standard-Backtrace-AC-Solutoin-short-and-clear

 

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2015-09-16 21:01  Grandyang  阅读(18306)  评论(6编辑  收藏  举报
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