如何在JavaScript的字符串数组中有效地找到包含另一个字符串的组字符串？

Question

For an array like:

["abc", "bc", "dd", "d", "ee", "ff", "e"] 

What would be an efficient way to get:

[["abc", "bc"],["dd", "d"],["ee", "e"]]

Explanation

["abc", "bc"] because "abc" contains "bc" 
["dd", "d"] because "dd" contains "d"
["ee", "e"] because "ee" contains "e"

Any new method including parallelism is also welcomed.

Answer 1

You can do it using reduce() . Check if the element is included by other element. If not then add it as key of accumulator. If its included then add it to that array. At last use Object.values() to get values(arrays). Use filter() to remove arrays having length = 1

 let arr = ["abc", "bc", "dd", "d", "ee", "ff", "e"] let res = Object.values(arr.reduce((ac,a,i) => { if(!arr.some((x,b) => x.includes(a) && i !== b)) ac[a] = [a]; else { for(let k in ac){ if(k.includes(a)){ ac[k].push(a) break; } } } return ac; },{})).filter(x => x.length -1) console.log(res) 

Answer 2

You can loop in reverse direction and using the function splice you can drop the element and push it into the current array.

 let arr = ["abc", "bc", "dd", "d", "ee", "ff", "e"]; let result = []; for (let i = arr.length - 1; i >= 0; i--) { let target = arr.splice(i, 1).pop(); if (target === undefined) continue; let current = [target], index = 0; while ((index = arr.findIndex(s => s.includes(target))) != -1) { current.push(arr[index]); arr.splice(index, 1); } if (current.length > 1) result.push(current); } console.log(result); 

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