为什么我的函数返回不同的数组？

Question

I have some code which is meant to do the following:

Given two arrays of strings a1 and a2 return a sorted array r in lexicographical order of the strings of a1 which are substrings of strings of a2 .
Arrays are written in "general" notation.

Now I'm doing a test where array1 = ["arp", "live", "strong"] and array2 = ["lively", "alive", "harp", "sharp", "armstrong"] . I think I've got it, but I don't understand why the function only returns the array ["arp", "strong"] when in the last for loop I replace newArray.splice(l, l+1) with newArray.splice(k, k+1) . Could anyone tell me why that is?

 function inArray(array1, array2) { var newArray = []; var sortedArray = []; for (var i in array2) { for (var j in array1) { if (array2[i].includes(array1[j])) { newArray.push(array1[j]); }; }; }; sortedArray = newArray.sort(); for (var k = 0; k < newArray.length; k++) { for (var l = 0; l < newArray.length; l++) { if (newArray[k] === newArray[l] && k != l) { newArray.splice(l, l + 1) } } } return sortedArray; }; console.log(inArray(["arp", "live", "strong"], ["lively", "alive", "harp", "sharp", "armstrong"])); 

Answer 1

The reason is that the second argument to splice() isn't the end position of the splice, but the count of elements that should be removed. So when k = 1 , you removed 2 elements, not just the element at index 1 .

Both newArray.splice(l, 1) and newArray.splice(k, 1) work correctly.

 function inArray(array1,array2){ var newArray = []; var sortedArray = []; for (var i in array2) { for (var j in array1) { if (array2[i].includes(array1[j])) { newArray.push(array1[j]); }; }; }; sortedArray = newArray.sort(); for (var k = 0; k < newArray.length; k++) { for (var l = 0; l < newArray.length; l++) { if (newArray[k] === newArray[l] && k != l) { newArray.splice(k, 1); } } } return sortedArray; }; console.log(inArray(["arp", "live", "strong"], ["lively", "alive", "harp", "sharp", "armstrong"])); 

Answer 2

In the comments already you get the answer. Use ES2015/ES6 always you can, benefit of it's adventages.

 const words1 = ["arp", "live", "strong"]; const words2 = ["lively", "alive", "harp", "sharp", "armstrong"]; const final = []; words1.forEach(v => { words2.forEach(v2 => { if (v2.includes(v)) { final.push(v); } }); }); const finalSorted = [...new Set(final.sort())]; // just to add html const pre = document.getElementById('result'); finalSorted.forEach(v => { let content = pre.textContent; content += `\\n- ${v}`; pre.textContent = content; }); 

 <label>Result:</label> <pre id="result"></pre> 

Answer 3

Looks like you're overcomplicating it a bit ;)

 a1 = ["arp", "live", "strong", "bazooka"] a2 = ["lively", "alive", "harp", "sharp", "armstrong"] result = a1 .filter(x => a2.some(y => y.includes(x))) .sort(); console.log(result);