Poker Quick Calculation 24 - Exhaustive (JavaScript)

Randomly draw 4 cards (no repetitions) A(1)2345678910J(11)Q(12)K(13) from the playing cards, please give an algorithm to make the result 24 through 4 operations, and each number is in the formula used once. If 24 cannot be obtained by the above rules, "24 cannot be calculated" is output.

If 4 numbers are given as 2 3 4 5, the output of the program should be an expression: (5+3-2)*4

Requirements: Submit algorithm ideas.
Ideas:
Let's first analyze how many cases are included in the exhaustive list:
1. Operands. Since it is a given 4 cards, all the 4 cards are arranged, a total of: 24 kinds
2. Operators. There are 3 operators between 4 numbers, each operator has 4 choices, total: 64
3. Operator precedence. Total: 5 kinds
According to the above statistics, there should be a total of: 24 64 5=7680 kinds

即如果用a、b、c、d代表这四个数，&代表某一种运算符，优先级依然用括号表示，那么只需要分别考虑以下这些形式：

 1、(a & b) & (c & d)
 2、((a & b) & c) & d
 3、(a & (b & c)) & d
 4、a & (b & (c & d))
 5、a & ((b & c) & d)

## problem solving code

const get24 = (str) => {
   let arr = str.split(' ')
   let transfer = (v) => {
       switch (v) {
           case 'A':
               return 1
           case 'J':
               return 11
           case 'Q':
               return 12
           case 'K':
               return 13
           default:
               return Number(v)
       }
   }
   //1.转换成真实数字
   let nums = []
   for (let i = 0; i < arr.length; i++) {
       nums[i] = transfer(arr[i])
   }
   //2.全排列-穷举不重复的排列-回溯
   let ans = [];
   let used = Array(nums.length).fill(false)
   let backTracing = (start, path) => {
       if (start === nums.length) {
           ans.push(path.slice())
           return
       }
       for (let i = 0; i < nums.length; ++i) {
           if (used[i] || (i > 0 && nums[i] === nums[i - 1] && !used[i - 1])) {
               continue;
           }
           path.push(nums[i])
           used[i] = true
           backTracing(start + 1, path)
           used[i] = false
           path.pop()
       }
   }
   nums.sort((a, b) => a - b)
   backTracing(0, [])
   /* 
   */
   //3.选择路径-穷举
   let dict = ['+', '-', '*', '/']
   for (let i = 0; i < ans.length; i++) {
       for (let a = 0; a < 4; a++) {
           for (let b = 0; b < 4; b++) {
               for (let c = 0; c < 4; c++) {
                   //(1-2)-3-4
                   let sum1 = eval(`(${ans[i][0]}${dict[a]}${ans[i][1]})${dict[b]}${ans[i][2]}${dict[c]}${ans[i][3]}`)
                   //1-(2-3)-4
                   let sum2 = eval(`${ans[i][0]}${dict[a]}(${ans[i][1]}${dict[b]}${ans[i][2]})${dict[c]}${ans[i][3]}`)
                   //1-2-(3-4)
                   let sum3 = eval(`${ans[i][0]}${dict[a]}${ans[i][1]}${dict[b]}(${ans[i][2]}${dict[c]}${ans[i][3]})`)
                   // (1-2-3)-4
                   let sum4 = eval(`(${ans[i][0]}${dict[a]}${ans[i][1]}${dict[b]}${ans[i][2]})${dict[c]}${ans[i][3]}`)
                   //1-(2-3-4)
                   let sum5 = eval(`${ans[i][0]}${dict[a]}(${ans[i][1]}${dict[b]}${ans[i][2]}${dict[c]}${ans[i][3]})`)
                   //(1-2)-(3-4)
                   let sum6 = eval(`(${ans[i][0]}${dict[a]}${ans[i][1]})${dict[b]}(${ans[i][2]}${dict[c]}${ans[i][3]})`)
                   // 1-((2-3)-4)
                   let sum7 = eval(`${ans[i][0]}${dict[a]}((${ans[i][1]}${dict[b]}${ans[i][2]})${dict[c]}${ans[i][3]})`)
                   //((1-2)-3)-4
                   let sum8 = eval(`((${ans[i][0]}${dict[a]}${ans[i][1]})${dict[b]}${ans[i][2]})${dict[c]}${ans[i][3]}`)
                   // 1-(2-(3-4))
                   let sum9 = eval(`${ans[i][0]}${dict[a]}(${ans[i][1]}${dict[b]}(${ans[i][2]}${dict[c]}${ans[i][3]}))`)
                   // 1-((2-3)-4)
                   let sum10 = eval(`${ans[i][0]}${dict[a]}((${ans[i][1]}${dict[b]}${ans[i][2]})${dict[c]}${ans[i][3]})`)
                   if ([sum1, sum2, sum3, sum4, sum5, sum6, sum7, sum8, sum9, sum10].includes(24)) {
                       return "YES"
                   }
               }
           }
       }
   }
   return "NO"
}

let str = 'A 2 3 6'
let str1 = '3 3 8 8'
console.log(get24(str))
console.log(get24(str1))