Question

这是我的代码：我想遍历数组的每一对元素，并且如果每对中的字符串包含一个或多个公共子字符串console.log(true)，否则为console.log(false)。因此，输出应为true，false，因为“ first”和“ criss”具有共同的子字符串（“ r”，“ i”，“ s”）这是我现在的代码；

const a = ["first", "hi"];
const b = ["criss", "student"];

function commonSubstring(a, b) {
  a.forEach(a1 =>
    b.forEach(a2 => {
      for (i = 0; i < a1.length; i++) {
        if (a1.charAt(i) == a2.charAt(i)) {
          console.log(true");
        }
      }
    })
  );
}
commonSubstring(a, b);

感谢提前答复！

Answer 1

您可以使用Set并检查字符是否常见。

function common(a, b) {
    return [...a].some(Set.prototype.has, new Set(b));
}

var a = ["first", "hi"],
    b = ["criss", "student"],
    result = a.map((v, i) => common(v, b[i]));

console.log(result);

Answer 2

const a = ["first", "hi"];
const b = ["criss", "student"];

function commonSubstring(a, b) {
  let result = [];
  a.forEach(a1 => {
    let found = false;
    b.forEach(a2 => {
      for (i = 0; i < a1.length; i++) {
        if (a1.charAt(i) == a2.charAt(i)) {
          //console.log(true);
          found = true;          
        }
      }
    })
    result.push(found);
    //console.log(found)
  });
 return result.join(',');
  
}
console.log(commonSubstring(a, b));

数组中的字符串包含常见的子字符串javascript

2 个答案: