From 458f94b8b94b6e9252dc23dc11f6627a4272e2b8 Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Thu, 1 Jun 2023 02:26:34 -0500
Subject: [PATCH] =?UTF-8?q?Update=200763.=E5=88=92=E5=88=86=E5=AD=97?=
 =?UTF-8?q?=E6=AF=8D=E5=8C=BA=E9=97=B4.md?=
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---
 problems/0763.划分字母区间.md | 111 +++++++++++-----------------
 1 file changed, 42 insertions(+), 69 deletions(-)

diff --git a/problems/0763.划分字母区间.md b/problems/0763.划分字母区间.md
index fbcafdc8..158f76d6 100644
--- a/problems/0763.划分字母区间.md
+++ b/problems/0763.划分字母区间.md
@@ -231,83 +231,56 @@ class Solution{
 ```
 
 ### Python 
+贪心（版本一）
 ```python
 class Solution:
     def partitionLabels(self, s: str) -> List[int]:
-        hash = [0] * 26
-        for i in range(len(s)):
-            hash[ord(s[i]) - ord('a')] = i
+        last_occurrence = {}  # 存储每个字符最后出现的位置
+        for i, ch in enumerate(s):
+            last_occurrence[ch] = i
+
         result = []
-        left = 0
-        right = 0
-        for i in range(len(s)):
-            right = max(right, hash[ord(s[i]) - ord('a')])
-            if i == right:
-                result.append(right - left + 1)
-                left = i + 1
+        start = 0
+        end = 0
+        for i, ch in enumerate(s):
+            end = max(end, last_occurrence[ch])  # 找到当前字符出现的最远位置
+            if i == end:  # 如果当前位置是最远位置，表示可以分割出一个区间
+                result.append(end - start + 1)
+                start = i + 1
+
         return result
-
-# 解法二（不相交区间法）
+         
+```
+贪心（版本二）与452.用最少数量的箭引爆气球 (opens new window)、435.无重叠区间 (opens new window)相同的思路。
+```python
 class Solution:
-    def partitionLabels(self, s: str) -> List[int]:
-        # 记录每个字母出现的区间
-        def getBord(s):
-            hash = [[-float('inf')] * 2 for _ in range(26)]
-            for i in range(len(s)):
-                if hash[ord(s[i]) - ord('a')][0] == -float('inf'):
-                    hash[ord(s[i]) - ord('a')][0] = i
-                hash[ord(s[i]) - ord('a')][1] = i
-            # 去除字符串中未出现的字母所占用区间
-            hash_filter = []
-            for item in hash:
-                if item[0] != -float('inf'): hash_filter.append(item)
-            return hash_filter
-
-        # 得到无重叠区间题意中的输入样例格式：区间列表
-        hash = getBord(s)
-        # 按照左边界从小到大排序
-        hash.sort(key= lambda x: x[0])
-        res = []
-        left = 0
-        # 记录最大右边界
-        right = hash[0][1]
+    def countLabels(self, s):
+        # 初始化一个长度为26的区间列表，初始值为负无穷
+        hash = [[float('-inf'), float('-inf')] for _ in range(26)]
+        hash_filter = []
+        for i in range(len(s)):
+            if hash[ord(s[i]) - ord('a')][0] == float('-inf'):
+                hash[ord(s[i]) - ord('a')][0] = i
+            hash[ord(s[i]) - ord('a')][1] = i
         for i in range(len(hash)):
-            # 一旦下一区间左边界大于当前右边界，即可认为出现分割点
-            if hash[i][0] > right:
-                res.append(right - left + 1)
-                left = hash[i][0]
-            # 实时更新最大右边界
-            right = max(right, hash[i][1])
-        # 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)
+            if hash[i][0] != float('-inf'):
+                hash_filter.append(hash[i])
+        return hash_filter
+
+    def partitionLabels(self, s):
+        res = []
+        hash = self.countLabels(s)
+        hash.sort(key=lambda x: x[0])  # 按左边界从小到大排序
+        rightBoard = hash[0][1]  # 记录最大右边界
+        leftBoard = 0
+        for i in range(1, len(hash)):
+            if hash[i][0] > rightBoard:  # 出现分割点
+                res.append(rightBoard - leftBoard + 1)
+                leftBoard = hash[i][0]
+            rightBoard = max(rightBoard, hash[i][1])
+        res.append(rightBoard - leftBoard + 1)  # 最右端
         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