今日学习
组合总和III
题目链接/文章讲解：https://programmercarl.com/0039.%E7%BB%84%E5%90%88%E6%80%BB%E5%92%8C.html
视频讲解：https://www.bilibili.com/video/BV1KT4y1M7HJ
第一想法
套模板，注意要在combinationSum3内重置result和sum
/**
* @param {number} k
* @param {number} n
* @return {number[][]}
*/
let sum = 0
let result = []
let path = []
var combinationSum3 = function (k, n) {
sum = 0
result = []
backtracking(k, n, 1)
return result
};
const backtracking = function (k, n, startindex) {
if (path.length === k) {
if (sum === n) {
result.push([...path])
return
}
}
for (let i = startindex; i <= 9; i++) {
if (sum + i > n) continue
path.push(i)
sum += i
backtracking(k, n, i + 1)
sum -= i
path.pop()
}
}
电话号码的字母组合
视频讲解：https://www.bilibili.com/video/BV1yV4y1V7Ug
第一想法
首先js中字符串是可以用const of遍历的，所以下文中的Number可以省略
/**
* @param {string} digits
* @return {string[]}
*/
const lettermap = ['', '', 'abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz']
let path = []
let result = []
var letterCombinations = function (digits) {
path = []
result = []
const k = digits.length
if (k === 0) {
return [];
}
backtracking(digits, k, 0)
return result
};
const backtracking = function (n, k, startindex) {
if (path.length === k) {
result.push(path.join(""))
return
}
for (const i of lettermap[Number(n[startindex])]) {
path.push(i)
backtracking(n, k, startindex + 1)
path.pop()
}
}
