From 2fc3e83bcf7c053420a5ebb485608c0ccce17ced Mon Sep 17 00:00:00 2001
From: ZerenZhang2022 <118794589+ZerenZhang2022@users.noreply.github.com>
Date: Sat, 18 Feb 2023 17:27:51 -0500
Subject: [PATCH] =?UTF-8?q?Update=200763.=E5=88=92=E5=88=86=E5=AD=97?=
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python 解法三：区间合并法 （结合下一题 56. Merge Intervals 的写法）
---
 problems/0763.划分字母区间.md | 27 +++++++++++++++++++++++++++
 1 file changed, 27 insertions(+)

diff --git a/problems/0763.划分字母区间.md b/problems/0763.划分字母区间.md
index 6c4f1c5e..6667e740 100644
--- a/problems/0763.划分字母区间.md
+++ b/problems/0763.划分字母区间.md
@@ -276,6 +276,33 @@ class Solution:
         # 最右侧区间（字符串长度为1时的特殊情况也包含于其中）
         res.append(right - left + 1)
         return res
+        
+# 解法三：区间合并法 （结合下一题 56. Merge Intervals 的写法）
+class Solution: # 
+    def partitionLabels(self, s: str) -> List[int]:
+        aaa = list(set(s))
+        #aaa.sort()
+        bbb = list(s)
+        ccc = []
+        for i in reversed(bbb):
+            ccc.append(i)
+        intervals = []
+        for i in range(len(aaa)):
+            intervals.append([bbb.index(aaa[i]),len(bbb)-ccc.index(aaa[i])-1])
+        # 先求出各个字母的存在区间，之后利用区间合并方法得出所有不相邻的最大区间。
+        intervals.sort(key = lambda x:x[0])
+        newinterval = []
+        left, right = intervals[0][0], intervals[0][1]
+        for i in range(1,len(intervals)):
+            if intervals[i][0] in range(left, right+1):
+                right = max(intervals[i][1],intervals[i-1][1],right)
+                left = min(intervals[i-1][0],left)
+            else: 
+                newinterval.append(right-left+1)
+                left = intervals[i][0]
+                right = intervals[i][1]
+        newinterval.append(right-left+1)
+        return newinterval                
 ```
 
 ### Go