题目：电话号码的字母组合
给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 任意顺序 返回。
给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number
//我的解答【有的方法写的还不健全 精力有限】
class LetterCombinations {
constructor() {
//字母数组
this.aryLetters = []
//数字数组
this.aryNumber = []
//组合后的数组
this.newData = []
//初始化
this.init()
}
//初始化
init() {
this.getLetters()
this.getNumber()
this.getNewData()
}
//获取字母数组
getLetters() {
for (var i = 0; i <= 25; i++) {
this.aryLetters.push(String.fromCharCode((65 + i)).toLowerCase());
}
}
//获取数字数组
getNumber() {
for (let i = 0; i <= 9; i++) {
this.aryNumber.push(i)
}
}
//组合数据【格式 {aryNumber[item]:letterAry】
//需要注意 0，1没有对应的字母；7 9 对应4个字母
//config配置参数【最后一个telNumber数字返回剩下的所有letterAry字母】
getNewData() {
let configParams = {
//截取字母长度
dafaultStrLength: 3,
//返回本身Number号
selfNum: [0, 1],
//特殊处理的Number[注意配置的长度不能大于默认长度dafaultStrLength]
otherNumber: [{
key: 7,
length: 4
}, {
key: 9,
length: 4
}]
}
let start = null
let end = null
//循环aryNumber每一项作为新数据对象中的key
this.aryNumber.forEach(i => {
//临时对象
let _obj = {}
//先过滤返回本身的aryNumber[i]
if (configParams.selfNum.includes(i)) {
_obj[i] = i
this.newData.push(_obj)
return
}
//默认长度配置
start = !end ? (i - 2) * configParams.dafaultStrLength : end
end = !end ? (i - 1) * configParams.dafaultStrLength : start + configParams.dafaultStrLength
//其他长度处理
configParams.otherNumber.forEach(j => {
if (i === j.key) {
end = !end ? (i - 1) * configParams.dafaultStrLength + Math.abs(configParams.dafaultStrLength - j.length) : start + configParams.dafaultStrLength + Math.abs(configParams.dafaultStrLength - j.length)
}
}
)
//截取字母作为aryNumber[i]的value
_obj[i] = this.aryLetters.slice(start, end)
//push
this.newData.push(_obj)
}
)
}
//获取需要组合的数组集
getAssociationData(...arg) {
if (arg.length === 0) {
console.log('请输入正确的数字格式')
return []
}
if (arg.includes(1) || arg.includes(0)) {
return []
}
//需要组合的数组集
let associationData = []
arg.forEach(k => {
if (k === 0 || k === 1) {
associationData.push([k])
return
}
associationData.push(this.newData.filter(m => m[k])[0][k])
}
)
return this.getFinishData(associationData)
}
//组合字母方法一
getFinishData(ary) {
let result = []
ary.forEach(i => {
let _ary = []
i.forEach(n => {
if (result.length > 0) {
_ary.push(...result.map(m => m + n))
} else {
_ary.push(n)
}
})
result = _ary
})
return result
}
}
let instandes = new LetterCombinations()
instandes.getAssociationData(2, 3, 4)
//二维数组字母组合方法二-递归
getFinishData(ary) {
if (!Array.isArray(ary[ary.length - 1])) {
return ary
}
let otherAry = ary.slice(2, initAry.length)
let thisAry = ary[0].map(item => ary[1].map(iten => item + iten)).flat()
let newAry = otherAry.length === 0 ? thisAry : [thisAry].concat(otherAry)
return dos(newAry)
}//二维数组字母组合方法三-递归
let formatUseDigits = [['a','b','c'],['d','e','f']]
let count = formatUseDigits.length
let result = []
function doOperating(str='', index=0) {
// 只push 符合长度的
if (str.length === count) {
result.push(str);
}
// 递归条件
if (index < count) {
const curr = formatUseDigits[index];
if (curr) {
for (const iterator of curr) {
// 拼接字符串
doOperating(str + iterator, index + 1);
}
}
}
}
doOperating()
console.log(result)
后端同事被卡住的一道面试题
急着给写答案，方法可能不是太精简
let n = 6, r = 10, x = [1, 7, 15, 20, 30, 50];
let res = []
function getSome(ary) {
if (ary.length == 0) {
return
} else if (ary.length == 1) {
res.push(ary[0])
return
}
let newAry = []
let a = ary[0]
//获取最后一个小于r的位置作为中间位置
let bAry = ary.filter(i => i <= ary[0] + r)
let b = ary[bAry.length - 1]
let c = ary[2]
if (b > a + r) {
newAry = ary.splice(1)
res.push(a)
getSome(newAry)
} else if (b <= a + r && !c) {
res.push(b)
} else if (b <= a + r && b + r <= c) {
newAry = ary.splice(bAry.length)
res.push(b)
getSome(newAry)
} else if (b <= a + r && b + r >= c) {
newAry = ary.splice(bAry.length + 1)
res.push(b)
getSome(newAry)
}
}
getSome(x)
console.log(`位置集合:${res}`,`摄像头数量:${res.length}`)
