leetcode -- Expression Add Operators --难点再看

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

主要参考https://leetcode.com/discuss/70597/clean-python-dfs-with-comments
其余参考都看不大懂。

这里
dfs() parameters:
num: remaining num string
temp: temporally string with operators added
cur: current result of “temp” string
last: last multiply-level number in “temp”. if next operator is “multiply”, “cur” and “last” will be updated。比如说temp = A+B, last = B; temp = A - B, last = -B; temp = A*B, last = A*B; temp = A+B*C, last = B*C.
res: result to return

思路就是枚举第一个操作数，然后在dfs里再枚举第二个操作数，然后在第一和第二个操作数中间枚举+-*.见code以及下面的图片

解释各种变量

解释如何dfs

对于上图第三点，我们要注意一个特殊情况，如果temp = A*B, 那么last = A*B, cur = A*B, 那么依旧是cur - last + last * int(val), 或者cur - old_last + new_last.

class Solution(object):    def addOperators(self, num, target):        res, self.target = [], target        for i in range(1,len(num)+1):            if i == 1 or (i > 1 and num[0] != "0"): # prevent "00*" as a number                self.dfs(num[i:], num[:i], int(num[:i]), int(num[:i]), res) # this step put first number in the string        return res    def dfs(self, num, temp, cur, last, res):        if not num:            if cur == self.target:                res.append(temp)            return        for i in range(1, len(num)+1):            val = num[:i]            if i == 1 or (i > 1 and num[0] != "0"): # prevent "00*" as a number                #注意这里是在temp和val中加操作符                self.dfs(num[i:], temp + "+" + val, cur+int(val), int(val), res)                self.dfs(num[i:], temp + "-" + val, cur-int(val), -int(val), res)                old_last = last                new_last = last*int(val)                self.dfs(num[i:], temp + "*" + val, cur-old_last+new_last, new_last, res)

参考：
http://bookshadow.com/weblog/2015/09/16/leetcode-expression-add-operators/

http://www.cnblogs.com/grandyang/p/4814506.html

http://blog.csdn.net/er_plough/article/details/48624117

http://nb4799.neu.edu/wordpress/?p=731、

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