[LeetCode][JavaScript]Different Ways to Add Parentheses

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 1

Input: "2-1-1".

((2-1)-1) = 0
(2-(1-1)) = 2

Output: [0, 2]


Example 2

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]

https://leetcode.com/problems/different-ways-to-add-parentheses/
 
 
 
 

 
 
按照符号分成左边和右边,分别计算左右的结果,结果是一个数组,其中包含了左边或右边的所有情况。
然后全排列这些组合,就是当前所有的结果。
举例来说,"2-1-1":
第一个"-"计算了2 - (1 - 1),
第二个"-"计算了(2 - 1) - 1。
 
 1 /**
 2  * @param {string} input
 3  * @return {number[]}
 4  */
 5 var diffWaysToCompute = function(input) {
 6     return compute(input);
 7 
 8     function compute(str){
 9         var res = [], i, j, k, left, right;
10         if(!/\+|\-|\*/.test(str)){ // + - * 
11             return [parseInt(str)];
12         }
13         for(i = 0 ; i < str.length; i++){
14             if(/\+|\-|\*/.test(str[i])){ // + - * 
15                 left = compute(str.substring(0, i));
16                 right = compute(str.substring(i + 1, str.length));
17                 for(j = 0; j < left.length; j++){
18                     for(k = 0; k < right.length; k++){
19                         if(str[i] === '+'){
20                             res.push(parseInt(left[j] + right[k]));
21                         }else if (str[i] === '-'){
22                             res.push(parseInt(left[j] - right[k]));
23                         }else if (str[i] === '*'){
24                             res.push(parseInt(left[j] * right[k]));
25                         }                       
26                     }
27                 }
28             }
29         }
30         return res;
31     }
32 };

 

 
 
 
posted @ 2015-09-20 17:23  `Liok  阅读(324)  评论(0编辑  收藏  举报