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Update 0763.划分字母区间.md

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problems/0763.划分字母区间.md
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@ -231,83 +231,56 @@ class Solution{
``` ```
### Python ### Python
贪心（版本一）
```python ```python
class Solution: class Solution:
def partitionLabels(self, s: str) -> List[int]: def partitionLabels(self, s: str) -> List[int]:
hash = [0] * 26 last_occurrence = {} # 存储每个字符最后出现的位置
for i in range(len(s)): for i, ch in enumerate(s):
hash[ord(s[i]) - ord('a')] = i last_occurrence[ch] = i
result = [] result = []
left = 0 start = 0
right = 0 end = 0
for i in range(len(s)): for i, ch in enumerate(s):
right = max(right, hash[ord(s[i]) - ord('a')]) end = max(end, last_occurrence[ch]) # 找到当前字符出现的最远位置
if i == right: 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(right - left + 1) result.append(end - start + 1)
left = i + 1 start = i + 1
return result 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: class Solution:
def partitionLabels(self, s: str) -> List[int]: def countLabels(self, s):
# 记录每个字母出现的区间 # 初始化一个长度为26的区间列表，初始值为负无穷
def getBord(s): hash = [[float('-inf'), float('-inf')] for _ in range(26)]
hash = [[-float('inf')] * 2 for _ in range(26)] hash_filter = []
for i in range(len(s)): for i in range(len(s)):
if hash[ord(s[i]) - ord('a')][0] == -float('inf'): if hash[ord(s[i]) - ord('a')][0] == float('-inf'):
hash[ord(s[i]) - ord('a')][0] = i hash[ord(s[i]) - ord('a')][0] = i
hash[ord(s[i]) - ord('a')][1] = 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]
for i in range(len(hash)): 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] != float('-inf'):
if hash[i][0] > right: hash_filter.append(hash[i])
res.append(right - left + 1) return hash_filter
left = hash[i][0]
# 实时更新最大右边界 def partitionLabels(self, s):
right = max(right, hash[i][1]) res = []
# 最右侧区间（字符串长度为1时的特殊情况也包含于其中） hash = self.countLabels(s)
res.append(right - left + 1) 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 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 ### Go