有没有更简单的方法来计算数组中行中的相同项目？

Question

This is my code

function repeat(arr){
    let string = (arr.join(''))
    let start = -1
    let count = 0
    while((start = string.indexOf('hhh', start + 1)) != -1 ){
        count += 1
        start += 3
    }
    while((start = string.indexOf('ttt', start + 1)) != -1 ){
        count += 1
        start += 3
    }
    console.log(count)
}

repeat(["h", "h", "h", "t", "h", "h", "t", "t", "t", "h", "t", "h", "h", "h", "h"])

In this example I tried to count how many times t or h goes 3 times in row.

Is there a way to do it without making a string? Or at least to merge block with while into one?

Answer 1

You can iterate each element in the array.

While doing so, you can keep track of:

• the total number of repeated groups that have occurred ( total )
• the value of the previous element seen in the array ( previousElement )
• the current number of elements that have repeated ( currentCount )

Below is an example of what I described above, including comments:

 function countRepeats (numOfRepeated, array) { let total = 0; let currentCount = 0; // Start with an object because no object strictly equals any other value: let previousElement = {}; for (const element of array) { if (element === previousElement) { // Increment the count: currentCount += 1; if (currentCount === numOfRepeated) { // Increment the total: total += 1; // Reset the count to 1: currentCount = 1; } } else { // Reset the count to 1: currentCount = 1; } // Update the previous element: previousElement = element; } return total; } const total = countRepeats( 3, ["h", "h", "h", "t", "h", "h", "t", "t", "t", "h", "t", "h", "h", "h", "h"], // 1 2 3 1 1 2 1 2 3 1 1 1 2 3 1 // 1 2 3 ); console.log(total); // 3