如何在一个数组中找到连续多少次相同的值？

Question

Hi I have question how to make algorithm to find how many in a row is the same value in array...

My array has 3 values: null, 0-0.99 and 1

Now I need to find in array how many times in a row is THE LAST value.

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My array = [null, null, null, 1, 0, 0, 1, 1, 1]

So I have three 1s as last items in an array how to count them in code?

Answer 1

If you only want to count how many continuous elements starting from the last one are equal, you can loop backwards like so:

 let arr = [null, null, null, 1, 0, 0, 1, 1, 1]; let last = arr[arr.length - 1]; let count = 1; for(let i = arr.length - 2; i >= 0; i--){ if(arr[i] === last){ ++count; } else break; } console.log(count);

If you want the length of the longest consecutive sequence of numbers equal to the last element of the array, you can loop through the whole array and reset the counter each time the element is not equal to the last one.

 let arr = [null, null, null, 1,1,1,1, 0, 0, 1, 1, 1]; let curr = 0; let maxLen = 1; let last = arr[arr.length - 1]; for(const elem of arr){ if(elem === last) maxLen = Math.max(maxLen, ++curr); else curr = 0; } console.log(maxLen);

Answer 2

 var myArray = [null, null, null, 1, 0, 0, 1, 1, 1]; var needle = myArray[myArray.length-1]; var cnt = 0; var oldCnt = 0; for(let i = 0;i<=myArray.length;i++){ if (myArray[i] == needle){ cnt++; }else{ if (cnt >= oldCnt)oldCnt=cnt; cnt = 0; } } console.log(oldCnt);