880

String/880.md

@@ -0,0 +1,38 @@
+## Decoded String at Index
+
+#### Description
+
+[link](https://leetcode.com/problems/decoded-string-at-index/)
+
+---
+
+#### Solution
+
+- See Code
+
+---
+
+#### Code
+
+O(n)
+
+```python
+class Solution:
+    '''
+    先求出刚好超过K的重复字符串长度，再往前寻找对应的第K个字符
+    当当前c为数字时，用N除去对应c，K同时也对N取余，等于在后续位置进行子问题查找
+    '''
+    def decodeAtIndex(self, S: str, K: int) -> str:
+        N = 0
+        for i, c in enumerate(S):
+            N = N * int(c) if c.isdigit() else N + 1
+            if K <= N: break
+        for j in range(i, -1, -1):
+            c = S[j]
+            if c.isdigit():
+                N /= int(c)
+                K %= N
+            else:
+                if K == N or K == 0: return c
+                N -= 1
+```