Question

让我说我有2个字符串：

var a = "abcdef", b = "abcdefgh";

我想找到第一个破坏完全匹配的索引，而不用遍历两个字符串并将每个字符与循环进行比较。

在此示例中，我想得到 6 ，因为 g 是破坏完整匹配的字符

Answer 1

使用reduce：

const findIndex = (a, b) => {
    if (b.length > a.length)
        return a.length;

    a = Array.from(a);
    b = Array.from(b);

    const result = a.reduce((acc, cur, ind) => {
        if (cur != b[ind] && (typeof acc.ind == 'undefined')) {
            acc.ind = ind;
        }
        return acc;
    }, {});

    return result.ind;
}

一个例子：

let a1 = "abcdef";
let b1 = "abcdefgh";

const findIndex = (a, b) => {
    if (b.length > a.length)
        return a.length;
        
    a = Array.from(a);
    b = Array.from(b);
    
    const result = a.reduce((acc, cur, ind) => {
        if (cur != b[ind] && (typeof acc.ind == 'undefined')) {
            acc.ind = ind;
        }
        return acc;
    }, {});

    return result.ind;
}

console.log(findIndex(a1, b1));
console.log(findIndex('aaa', ''));
console.log(findIndex('hey:)', 'hey!'));

此外，带有for循环的版本。特别感谢@ Andreas关于如何避免if语句的建议：

const findIndex = (a, b) => {
    for (let i = 0, l = Math.max(a.length, b.length); i < l; i++) {
        if (a[i] != b[i])
            return i;
     }
     return 'equal';
}

一个例子：

let a = "abcdef";
let b = "abcdefgh";

const findIndex = (a, b) => {
    for (let i = 0, l = Math.max(a.length, b.length); i < l; i++) {
        if (a[i] != b[i])
            return i;
    }
    return 'equal';
}

console.log(findIndex(a, b));
console.log(findIndex('aaa', ''));
console.log(findIndex('hey:)', 'hey!'));

Answer 2

尝试一下-var index = b.indexOf(a) + a.length

var a = "abcdef", b = "abcdefgh";
document.write(b.indexOf(a) + a.length);

Answer 3

在这种情况下，请尝试a.lastIndexOf(b.charAt(b.length));

var a = "abcdef", b = "abcdefgh";
document.write(a.lastIndexOf(b.charAt(b.length)));

Answer 4

这是没有循环的更快解决方案-

var stringA = "abcdef",
  stringB = "abcdefgh",
  firstOccurrence = stringB.indexOf(stringA);

if (firstOccurrence === -1) {
  alert('Search string Not found');
} else {
  var stringALength = stringA.length;
  document.write(firstOccurrence + stringALength);
}

查找破坏两个字符串之间匹配的索引

4 个答案: