400

Author: ProHiryu

Math/400.md

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+## 400 Nth Digit
+
+#### Description
+
+[link](https://leetcode.com/problems/nth-digit/)
+
+---
+
+#### Solution
+
+observation: 假设我们要找第2886 个位 (就target数 是998，target bit 是8)
+
+- 2886 - 9 - 9 * 10 * 2 = 2697 < 9*10*10*3 = 2700
+- target 就落在了区域3中( 100- 999 )，因为这个区域中每一位都有3位，所以可以计算target
+- target 数是以 100 为起始数，(2697 - 1)/3 = 898 为100以后的数，-1是因为起始数本来就算一个数，不能重复计算
+- target 数 = 100 + 898 = 998
+- (2697-1) % 3 = 2 就是 998 的target bit
+- target bit = 998.charAt( 2 ) = 8;
+
+---
+
+#### Code
+
+<!-- O(n) -->
+
+```python
+class Solution:
+    def findNthDigit(self, n: int) -> int:
+        start = 1
+        base = 9
+        digits = 1
+        
+        # step 1 find how many digits the number need
+        while n > base * digits:
+            n -= base * digits
+            digits += 1
+            base *= 10
+            start *= 10
+        
+        # step 2 find what number we need
+        target = start + (n - 1) // digits
+        reminder = (n - 1) % digits
+        
+        # step 3 find which digit we need
+        return int(str(target)[reminder])
+```