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🐱(backtrack): 357. 计算各个位数不同的数字个数

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`@@ -168,4 +168,78 @@ func dfs(ans *[][]int, element []int, start int, num int, k int) {`
`168`	`168`	`}`
`169`	`169`	```
`170`	`170`
`171`		`-<!-- tabs:end -->`
	`171`	`+<!-- tabs:end -->`
	`172`	`+`
	`173`	`+## 357. 计算各个位数不同的数字个数`
	`174`	`+`
	`175`	`+[原题链接](https://leetcode-cn.com/problems/count-numbers-with-unique-digits/)`
	`176`	`+`
	`177`	`+### 解法一:回溯`
	`178`	`+`
	`179`	+用 `tags` 数组标记 0~9 数字出现的次数,再调用递归后进行回溯。注意对 0 进行特殊处理。
	`180`	`+`
	`181`	+```python
	`182`	`+class Solution:`
	`183`	`+ ans = 0`
	`184`	`+ tags = [] # 记录数字出现的次数`
	`185`	`+ def countNumbersWithUniqueDigits(self, n: int) -> int:`
	`186`	`+ # 0-9 的数字`
	`187`	`+ self.tags = [0 for _ in range(10)]`
	`188`	`+ # 枚举所有数字:回溯`
	`189`	`+ """`
	`190`	`+ r: 轮次`
	`191`	`+ num: 数字`
	`192`	`+ """`
	`193`	`+ def dfs(r, num = 0):`
	`194`	`+ if r <= 0:`
	`195`	`+ # 结束递归`
	`196`	`+ return`
	`197`	`+`
	`198`	`+ for i in range(10):`
	`199`	`+ # 循环 0-9`
	`200`	`+ if num % 10 != i and self.tags[i] == 0:`
	`201`	`+ # 条件枝剪`
	`202`	`+ self.ans += 1`
	`203`	`+ # 数字出现标记`
	`204`	`+ self.tags[i] = 1`
	`205`	`+ dfs(r - 1, num * 10 + i)`
	`206`	`+ # 回溯`
	`207`	`+ self.tags[i] = 0`
	`208`	`+`
	`209`	`+ dfs(n)`
	`210`	`+ # 补充 0`
	`211`	`+ return self.ans + 1`
	`212`	+```
	`213`	`+`
	`214`	`+### 解法二:动态规划`
	`215`	`+`
	`216`	`+排列组合。`
	`217`	`+`
	`218`	+- `f(0) = 1`
	`219`	+- `f(1) = 9`
	`220`	+- `f(2) = 9 * 9 + f(1)`
	`221`	`+ - 第一个数字选择 1~9`
	`222`	`+ - 第二个数在 0~9 中选择和第一个数不同的数`
	`223`	+- `f(3) = 9 * 9 * 8 + f(2)`
	`224`	`+`
	`225`	`+可以推出动态规划方程:`
	`226`	`+`
	`227`	+```
	`228`	`+dp[i] = sum + dp[i - 1]`
	`229`	+```
	`230`	`+`
	`231`	+```python
	`232`	`+class Solution:`
	`233`	`+ def countNumbersWithUniqueDigits(self, n: int) -> int:`
	`234`	`+ dp = [0 for _ in range(n + 1)]`
	`235`	`+ # n = 1 时`
	`236`	`+ dp[0] = 1`
	`237`	`+`
	`238`	`+ for i in range(1, n + 1):`
	`239`	`+ s = 9`
	`240`	`+ for j in range(1, i):`
	`241`	`+ s *= (10 - j)`
	`242`	`+ dp[i] = s + dp[i - 1]`
	`243`	`+`
	`244`	`+ return dp[n]`
	`245`	+```

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