数组字符串的字谜
Anagram of strings of arrays
这个程序只能处理一个单词,但我想传递一组字符串(多个单词)。
let output = getAnagrams("CATCH"); //it is working for this
let output = getAnagrams(["catch", "Priest", "Monkey", "Bruise"]);
我希望它能为此工作。
function swap(chars, i, j) {
var tmp = chars[i];
chars[i] = chars[j];
chars[j] = tmp;
}
function getAnagrams(input) {
let newInput = input.toString().toLowerCase();
console.log(newInput);
var counter = [],
anagrams = [],
chars = newInput.split(''),
length = chars.length,
i;
for (i = 0; i < length; i++) {
counter[i] = 0;
}
anagrams.push(newInput);
i = 0;
while (i < length) {
if (counter[i] < i) {
swap(chars, i % 2 === 1 ? Counter[i] : 0, i);
counter[i]++;
i = 0;
anagrams.push(chars.join(''));
} else {
counter[i] = 0;
i++;
}
}
// return anagrams;
}
因为您已经有了一个接受 1 个字符串的方法,为什么不为数组中的每个字符串调用它,然后使用 flatMap
展平返回的数组
function getAnagrams(input) {
let newInput = input.toString().toLowerCase();
var counter = [],
anagrams = [],
chars = newInput.split(''),
length = chars.length,
i;
for (i = 0; i < length; i++) {
counter[i] = 0;
}
anagrams.push(newInput);
i = 0;
while (i < length) {
if (counter[i] < i) {
swap(chars, i % 2 === 1 ? counter[i] : 0, i);
counter[i]++;
i = 0;
anagrams.push(chars.join(''));
} else {
counter[i] = 0;
i++;
}
}
return anagrams;
}
function swap(arr,i,j){
const tmp = arr[i];
arr[i] = arr[j]
arr[j] = tmp
}
const result = ["catch", "Priest", "Monkey", "Bruise"].flatMap(i => getAnagrams(i))
console.log(result)
这个程序只能处理一个单词,但我想传递一组字符串(多个单词)。
let output = getAnagrams("CATCH"); //it is working for this
let output = getAnagrams(["catch", "Priest", "Monkey", "Bruise"]);
我希望它能为此工作。
function swap(chars, i, j) {
var tmp = chars[i];
chars[i] = chars[j];
chars[j] = tmp;
}
function getAnagrams(input) {
let newInput = input.toString().toLowerCase();
console.log(newInput);
var counter = [],
anagrams = [],
chars = newInput.split(''),
length = chars.length,
i;
for (i = 0; i < length; i++) {
counter[i] = 0;
}
anagrams.push(newInput);
i = 0;
while (i < length) {
if (counter[i] < i) {
swap(chars, i % 2 === 1 ? Counter[i] : 0, i);
counter[i]++;
i = 0;
anagrams.push(chars.join(''));
} else {
counter[i] = 0;
i++;
}
}
// return anagrams;
}
因为您已经有了一个接受 1 个字符串的方法,为什么不为数组中的每个字符串调用它,然后使用 flatMap
function getAnagrams(input) {
let newInput = input.toString().toLowerCase();
var counter = [],
anagrams = [],
chars = newInput.split(''),
length = chars.length,
i;
for (i = 0; i < length; i++) {
counter[i] = 0;
}
anagrams.push(newInput);
i = 0;
while (i < length) {
if (counter[i] < i) {
swap(chars, i % 2 === 1 ? counter[i] : 0, i);
counter[i]++;
i = 0;
anagrams.push(chars.join(''));
} else {
counter[i] = 0;
i++;
}
}
return anagrams;
}
function swap(arr,i,j){
const tmp = arr[i];
arr[i] = arr[j]
arr[j] = tmp
}
const result = ["catch", "Priest", "Monkey", "Bruise"].flatMap(i => getAnagrams(i))
console.log(result)