【LeetCode 241】Different Ways to Add Parentheses

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are+, - and *.

Example ：

Input: "2*3-4*5"

(2*(3-(4*5))) = -34

((2*3)-(4*5)) = -14

((2*(3-4))*5) = -10

(2*((3-4)*5)) = -10

(((2*3)-4)*5) = 10

Output: [-34, -14, -10, -10, 10]

题意：

　　给一串表达式（由+、-、*以及数字组成），通过加括号改变运算顺序，求所有可能的结果。

思路：

　　没有想到什么比较巧妙的方法，只能递归所有的可能咯，其中有很多重复计算，等找到好办法了再来更新吧。

C++：

 1 class Solution {

 2 public:

 3 

 4     //string转为int

 5     int getInts(string str)

 6     {

 7         int ret = 0;

 8 

 9         stringstream ss;

10         ss << str;

11         ss >> ret;

12 

13         return ret;

14     }

15 

16     vector<int> rec(vector<int>& num, int start, int ends)

17     {

18         vector<int> vec;

19     

20         if(start == ends)

21         {

22             vec.push_back(num[start]);

23             return vec;

24         }

25     

26         for(int i = start; i <= ends; i++)

27         {

28             if(num[i] == -3)

29             {

30                 vector<int> vecl = rec(num, start, i - 1);

31                 vector<int> vecr = rec(num, i + 1, ends);

32     

33                 vector<int>::iterator itl = vecl.begin();

34                 for(; itl != vecl.end(); itl++)

35                 {

36                     vector<int>::iterator itr = vecr.begin();

37                     for(; itr != vecr.end(); itr++)

38                         vec.push_back((*itl) + (*itr));

39                 }

40             }

41             if(num[i] == -1)

42             {

43                 vector<int> vecl = rec(num, start, i - 1);

44                 vector<int> vecr = rec(num, i + 1, ends);

45     

46                 vector<int>::iterator itl = vecl.begin();

47                 for(; itl != vecl.end(); itl++)

48                 {

49                     vector<int>::iterator itr = vecr.begin();

50                     for(; itr != vecr.end(); itr++)

51                         vec.push_back((*itl) - (*itr));

52                 }

53             }

54             if(num[i] == -4)

55             {

56                 vector<int> vecl = rec(num, start, i - 1);

57                 vector<int> vecr = rec(num, i + 1, ends);

58     

59                 vector<int>::iterator itl = vecl.begin();

60                 for(; itl != vecl.end(); itl++)

61                 {

62                     vector<int>::iterator itr = vecr.begin();

63                     for(; itr != vecr.end(); itr++)

64                         vec.push_back((*itl) * (*itr));

65                 }

66             }

67         }

68         return vec;

69     }

70     

71     vector<int> diffWaysToCompute(string input) {

72         

73         //先把字符串转换为一个int型的数组，符号用负数来表示

74         vector<int> nums;

75         int index = 0;

76         for(int i = 0; i < input.size(); i++)

77         {

78             if(input[i] == '+' || input[i] == '-' || input[i] == '*')

79             {

80                 string str = input.substr(index, i - index);

81                 nums.push_back(getInts(str));

82                 nums.push_back(input[i]-46); //加号转换为-3， 减号转换为-1，乘号为-4

83                 index = i + 1;

84             }

85             if(i == input.size() - 1)

86             {

87                 string str = input.substr(index, input.size() - index);

88                 nums.push_back(getInts(str));

89             }

90         }

91         

92         int len = nums.size();

93         if(len == 0 || len == 1)

94             return nums;

95         

96         vector<int> ret = rec(nums, 0, len - 1);

97         return ret;

98     }

99 };

AC： https://leetcode.com/submissions/detail/34357736/