Create 763._Partition_Labels.md

Author: 0xkookoo

docs/Leetcode_Solutions/Python/763._Partition_Labels.md

@@ -0,0 +1,76 @@
+# 763. Partition Labels
+
+**<font color=red>难度: Medium</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/partition-labels/description/
+
+> 内容描述
+
+```
+
+A string S of lowercase letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and return a list of integers representing the size of these parts.
+
+Example 1:
+Input: S = "ababcbacadefegdehijhklij"
+Output: [9,7,8]
+Explanation:
+The partition is "ababcbaca", "defegde", "hijhklij".
+This is a partition so that each letter appears in at most one part.
+A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits S into less parts.
+Note:
+
+S will have length in range [1, 500].
+S will consist of lowercase letters ('a' to 'z') only.
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(N)******- 空间复杂度: O(1)******
+
+
+从第一个字母开始，只要我们包含了这个字母，那么它在S中出现的最后一次我们也要包含进去，当前出现过的所有的字母，我们每次都要更新最后出现的那一次，
+并且要一直遍历到其最后一次，这样我们可以维护一个当前至少需要遍历到的end，然后只要出现一个新的字母，
+就更新一个这个end（取新字母最后出现idx和end之间的大者），最后，当我们遍历到一个idx的时候，如果这个idx与我们的end相等，
+说明截止当前所有出现的字母都在我们的[start, end]字串当中了，直接append它的长度到res中，一直遍历到最后，返回res
+
+beats 98.39%
+
+```python
+class Solution(object):
+    def partitionLabels(self, S):
+        """
+        :type S: str
+        :rtype: List[int]
+        """
+        if not S or len(S) == 0:
+            return True
+        lookup = {}
+        for i, c in enumerate(S):
+            lookup[c] = i
+        start, end, res = 0, 0, []
+        for i, c in enumerate(S):
+            end = max(end, lookup[c])
+            if i == end:
+                res.append(end-start+1)
+                start = i + 1
+        return res
+```
+
+
+
+
+
+
+
+
+
+
+
+
+
+