我需要一个函数来搜索数组并返回最后一个字符串后的最后一组数字？

Question

I need a function call getLastNumbers which searches through an array and returns the last set of numbers after the last operator.

for example:

[1, 1, 1, 1, 1, 1, "+", 2, 2, 2, 2, 2, 2, "*", 3, 3, 3, 3, 3, "-", 6, 6, 6, 6] Returns 6666

[1, 1, 1, 1, "+", 6, 6, 6, 6, 6, 6, "*", 9, 9, 9, 9, 9, "-", 6, 6, 6, 6, "+", 3, 3, 3] Returns 333

[7, "*", 7, "-", 6, "+", 3, "+", 6, "+", 3, 3, 3, 3, 3, "+", 6, "-", 5, 5, 5, 5, 5, 5, "*", 9, 9] Returns 99

[5, "*", 5] returns 5

To clarify, I will never know how many numbers or operators the user may input.

See my long winded attempt below attempt below:

function getLastNumber(arr){
let number = arr.filter(i => i === '*' || i === '/' || i === '+' || i === '-' ).length;

  let counter = 0;
  let arrayLength = arr.length;
  let lastOpIndex = [];

 for(let i = 0; i < arr.length; i++) {
   if(arr[i] === '*' || arr[i] === '+' ||arr[i] === '-' ||arr[i] === '/'){
     counter++;
     if (counter === number) {
       lastOpIndex.push(arr.indexOf(arr[i]));
     }
   }
 }

  let lastOpNum = parseInt(lastOpIndex.toString());
  let startNum = lastOpNum + 1;
  let mathNum = (arrayLength - lastOpNum) - 1;


  let finalArray = [];

  for(let i = startNum; i < arr.length; i++) {
    finalArray.push(arr[i]);
  }

  let strResult = finalArray.toString().split(',').join('');

  let finalResult = parseInt(strResult);

  return finalResult;


}



getLastNumber([1, "+", 3, 3, 3, 3, 3, "+", 6, "+", 9, "*", 6, 6, 6, 3, 6, 6, 6, "-", 5, "*", 9, 9, 9, 9, 9, 9, 9, '/', 6, 8, 9, 9, '*', 8, 8, 8, 7]);

Answer 1

As the question is open again i will extend on my comment. I think they way you're trying to achieve this is a bit to complicated.

Here is my approach just turning the problem arround and starting with the last index instead of the first one. As soon as I find an operator I just break the loop and return the saved values in reversed order.

 function getLastNumber(arr){ let ret = []; for(let i = arr.length - 1; i >= 0 ; i--) { if(arr[i] !== '*' && arr[i] !== '+' && arr[i] !== '-' && arr[i] !== '/') { ret.push(arr[i]); } else { break; } } return ret.reverse().toString(); } console.log(getLastNumber([1, "+", 3, 3, 3, 3, 3, "+", 6, "+", 9, "*", 6, 6, 6, 3, 6, 6, 6, "-", 5, "*", 9, 9, 9, 9, 9, 9, 9, '/', 6, 8, 9, 9, '*', 8, 8, 8, 7])); console.log(getLastNumber([1, 1, 1, 1, "+", 6, 6, 6, 6, 6, 6, "*", 9, 9, 9, 9, 9, "-", 6, 6, 6, 6, "+", 3, 3, 3])); console.log(getLastNumber([7, "*", 7, "-", 6, "+", 3, "+", 6, "+", 3, 3, 3, 3, 3, "+", 6, "-", 5, 5, 5, 5, 5, 5, "*", 9, 9])); console.log(getLastNumber([5, "*", 5]))

Answer 2

 function filterNumbers(arr){ let a = arr.lastIndexOf("*"); // find last index of * let b = arr.lastIndexOf("+"); // find last index of + let c = arr.lastIndexOf("-"); // find last index of - let d = arr.lastIndexOf("/"); // find last index of / let ind = Math.max(a, b, c, d); //find the max index console.log(ind); let newarr = arr.splice(ind+1,arr.length); //splice array console.log(newarr); return newarr; } filterNumbers([1, 1, 1, 1, 1, 1, "+", 2, 2, 2, 2, 2, 2, "*", 3, 3, 3, 3, 3, 6, 6, 6, 6,1]); filterNumbers([1, 1, 1, 1, 1, 1, "+", 2, 2, 2, 2, 2, 2, "*", 3, 3, 3, 3,"/", 3, 6, 6, 6, 6,1]);