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Python LeetCode（七）
不会做
1. Search in Rotated Sorted Array
class Solution(object): def search(self, nums, target): """ :type nums: List[int] :type target: int :rtype: int """ if nums ==[]: return -1 left, right = 0, len(nums)-1 infinity = 2**31 mid = int(round((left+right)/2)) while left < right: mid = int(round((left+right)/2)) # 如果为true，则target和mid都在右边，否则都在左边 if (nums[0] > target) == (nums[0] > nums[mid]): number = nums[mid] # 同个方向，按照二分来 elif nums[0] > target: # target在右边，mid在左边 number = -infinity # 保证left向右移动 else: # target在左边，mid在右边 number = infinity # 保证right向左移动 if number < target: left = mid+1 else: right = mid number = 0 if nums != [] and nums[left] == target: return left else: return -1
class Solution(object): def search(self, nums, target): if not nums: return -1 lo, hi = 0, len(nums) - 1 while lo < hi: mid = (lo + hi) / 2 #python2中返回的是int，但是python3中则会返回float if nums[mid] == target: return mid if nums[lo] <= nums[mid]: # sorted order if target>=nums[lo] and target<nums[mid]: hi = mid - 1 else: lo = mid + 1 else: # drop between low, high if target>nums[mid] and target<=nums[hi]: lo = mid + 1 else: hi = mid - 1 return lo if nums[lo] == target else -1
from bisect import bisect_left def binary_search(nums, target): pos = bisect_left(nums, target) if nums[min(pos, len(nums) - 1)] == target: return pos else: return -1 class Solution(object): def search(self, nums, target): """ :type nums: List[int] :type target: int :rtype: int """ # Empty array if not nums: return -1 # Size 1 if len(nums) == 1: return 0 if nums[0] == target else -1 # Array is sorted if nums[-1] > nums[0]: return binary_search(nums, target) # Search left m = len(nums) // 2 pos = self.search(nums[:m], target) if pos > -1: return pos # Search right pos = self.search(nums[m:], target) if pos == -1: return pos else: return pos + m
2. Search in Rotated Sorted Array II
class Solution(object): def search(self, nums, target): """ :type nums: List[int] :type target: int :rtype: bool """ left, right = 0, len(nums)-1 mid = left while left <= right: mid = int(round((left + right)/2)) if nums[mid] == target: return True if nums[mid] > nums[left]: # 左边部分有序 if nums[left] <= target < nums[mid]: # target 在左边 right = mid-1 else: left = mid+1 elif nums[mid] < nums[left]: # 右边部分有序 if nums[mid] < target <= nums[right]: # target 在右边 left = mid+1 else: right = mid-1 elif nums[mid] == nums[left]: # 特殊情况 left+=1 return False
class Solution(object): def search(self, nums, target): """ :type nums: List[int] :type target: int :rtype: bool """ left, right = 0, len(nums)-1 mid = left while left <= right: mid = int(round((left + right)/2)) if nums[mid] == target: return True if nums[mid] > nums[right]: # 左边部分有序 if nums[left] <= target < nums[mid]: # target 在左边 right = mid-1 else: left = mid+1 elif nums[mid] < nums[right]: # 右边部分有序 if nums[mid] < target <= nums[right]: # target 在右边 left = mid+1 else: right = mid-1 elif nums[mid] == nums[right]: # 特殊情况 right-=1 return False
class Solution(object): def search(self, nums, target): """ :type nums: List[int] :type target: int :rtype: bool """ if not nums or target is None: return False return self.helper(nums, 0, len(nums) - 1, target) def helper(self, nums, left, right, target): # base case if left > right: return False mid = left + (right - left) / 2 if nums[mid] == target: return True if nums[mid] > nums[left]: # means the left part is sorted. if nums[left] <= target < nums[mid]: return self.binary_search(nums, left, mid - 1, target) # recursion on right part. return self.helper(nums, mid + 1, right, target) elif nums[mid] < nums[left]: # means the right part is sorted. if nums[mid] < target <= nums[right]: return self.binary_search(nums, mid + 1, right, target) # recursion on left part. return self.helper(nums, left, mid - 1, target) else: # recursion on both part. return self.helper(nums, left, mid - 1, target) or self.helper(nums, mid + 1, right, target) def binary_search(self, nums, left, right, target): while left <= right: mid = left + (right - left) / 2 if nums[mid] == target: return True elif nums[mid] < target: left = mid + 1 else: right = mid - 1 return False
3. Wildcard Matching
class Solution(object): # p为匹配模式，s为字符串 def recursive(self, s, p, si, pi, cur): first = True n_cur = cur while si < len(s) and pi < len(p) and (s[si] == p[pi] or p[pi] == ‘?‘): si += 1 pi += 1 if pi == len(p): return si == len(s) if p[pi] == ‘*‘: while pi < len(p) and p[pi] == ‘*‘: pi += 1 if pi >= len(p): return True for i in range(si, len(s)): # 表明开始重合，从这里再度开始递归 if p[pi] != s[i] and p[pi] != ‘?‘: continue if first: cur += 1 first = False # 可能存在多次重合但是还不算真正匹配的情况 if self.recursive(s, p, i, pi, cur + 1): return True if cur > n_cur + 1: # 正常来说n_cur = cur + 1 return False return False def isMatch(self, s, p): """ :type s: str :type p: str :rtype: bool """ return self.recursive(s, p, 0, 0, 0) class Solution(object): # p为匹配模式，s为字符串 def recursive(self, s, p, si, pi): while si < len(s) and pi < len(p) and (s[si] == p[pi] or p[pi] == ‘?‘): si += 1 pi += 1 if pi == len(p): return si == len(s) if p[pi] == ‘*‘: while pi < len(p) and p[pi] == ‘*‘: pi += 1 if pi >= len(p): return True for i in range(si, len(s)): # 表明开始重合，从这里再度开始递归 if p[pi] != s[i] and p[pi] != ‘?‘: continue # 可能存在多次重合但是还不算真正匹配的情况 if self.recursive(s, p, i, pi): return True return False def isMatch(self, s, p): """ :type s: str :type p: str :rtype: bool """ return self.recursive(s, p, 0, 0)
pattern = "a*bc"
str = "abbc"| \ | \ | a | b | b | c |
|---|---|---|---|---|---|
| \ | T | F | F | F | F |
| a | F | T | F | F | F |
| * | F | T | T | T | T |
| b | F | F | T | T | F |
| c | F | F | F | F | T |
- dp[i][j] = dp[i-1]j-1
- dp[i][j] = dp[i-1][j] or dp[i]j-1
- 其它情况为False
class Solution(object): # p为匹配模式，s为字符串 def isMatch(self, s, p): """ :type s: str :type p: str :rtype: bool """ if len(s) != len(p) - p.count(‘*‘): return False newp = "" i = 0 while i < len(p): newp += p[i] if p[i] == ‘*‘: while i + 1 < len(p) and p[i + 1] == ‘*‘: i += 1 i += 1 sl, pl = len(s), len(newp) dp = [[False for x in range(pl + 1)] for y in range(sl + 1)] dp[0][0] = True if pl > 0 and p[0] == ‘*‘: dp[0][1] = True for x in range(1, sl + 1): for y in range(1, pl + 1): if newp[y - 1] != ‘*‘: dp[x][y] = dp[x - 1][y - 1] and (s[x - 1] == newp[y - 1] or newp[y - 1] == ‘?‘) else: dp[x][y] = dp[x - 1][y] or dp[x][y - 1] return dp[sl][pl]
class Solution: # @return a boolean def isMatch(self, s, p): length = len(s) if len(p) - p.count(‘*‘) > length: return False dp = [True] + [False]*length for i in p: if i != ‘*‘: # 因为依赖项是前面的值，所以不能从前面往后面扫，得从后往前计算 for n in reversed(range(length)): dp[n+1] = dp[n] and (i == s[n] or i == ‘?‘) else: # 更新当前行的数据 for n in range(1, length+1): dp[n] = dp[n-1] or dp[n] dp[0] = dp[0] and i == ‘*‘ return dp[-1]
贪心算法
| 下标 | 描述 |
|---|---|
| si | 待匹配字符串的移动下标 |
| pi | 模式串的移动下标 |
| lastmatch | 上一次匹配的待匹配字符串的下标 |
| laststar | 上一次匹配的模式串的下标 |
- 如果当前相等或者模式串中字符为
?，则移动相互的下标即可； - 如果当前模式串字符为
*，分别纪录lastmatch、laststar，并且移动模式串下标，但是不移动待匹配字符串下标，因为可能存在匹配0个字符串的情况； - 如果当前相互对应的字符不再相等且不为
*，如果前面有*号，说明之前的匹配失败了，模式字符串下标回到之前纪录laststar的后一位，不再移动，专门用来给待匹配字符串字符来匹配，这段时间内，si会不断的向前移动，直到匹配到相互的值相等才移动模式字符串的下标； - 如果前面的情况都不符合，则肯定为False；
class Solution(object): # p为匹配模式，s为字符串 def isMatch(self, s, p): si, pi = 0, 0 lastmatch, laststar = -1, -1 sl, pl = len(s), len(p) if pl - p.count(‘*‘) > sl: return False # 注意条件顺序 while si < sl: if pi < pl and (s[si] == p[pi] or p[pi] == ‘?‘): pi += 1 si += 1 elif pi < pl and p[pi] == ‘*‘: lastmatch, laststar = si, pi # 之所以不更新lastmatch是因为考虑到*只匹配0个字符串 pi += 1 # 再次进到这个判断，说明当前下标对应的值不相等 elif laststar != -1: pi = laststar + 1 # pi当前不是*，并且回到上一次星的后面，专门用来给si匹配 lastmatch += 1 # 必须更新lastmatch，因为之前已经不想等，如果在回到开始的状态就会陷入死循环 si = lastmatch else: return False # 可能存在p的末尾都是*的情况 while pi < len(p) and p[pi] == ‘*‘: pi += 1 # 最后匹配成功模式字符串的下标必然为其长度，表示已经匹配完成 return pi == pl
class Solution(object): def isMatch(self, s, p): """ :type s: str :type p: str :rtype: bool """ seen = {} wild_single, wild_multi = "?", "*" # seen has the pattern - source tuple as key, and bool result as success source, pattern = s, p def is_match(sindex, pindex): key = (sindex, pindex) if key in seen: return seen[key] result = True # if there‘s no string, and pattern is not only * then fail if sindex >= len(source): for wildindex in xrange(pindex, len(pattern)): if pattern[wildindex] != wild_multi: result = False break # there‘s a string, but no pattern elif pindex >= len(pattern): result = False # if next pattern is multi though, that‘s something elif pattern[pindex] == wild_multi: # for zero, simply check sindex, pindex + 1 result = is_match(sindex, pindex + 1) # just for easier debug # if zero, than it‘s a match # otherwise we need to check multi # for that, if char is not a wild, then it has to match the source, result = result or is_match(sindex + 1, pindex) else: # either a regular char, or wild_single result = (( pattern[pindex] == wild_single or pattern[pindex] == source[sindex]) and is_match(sindex + 1, pindex + 1)) seen[key] = result return result if (len(p) - p.count(wild_multi) > len(s)): return False return is_match(0, 0)
不同的做法
1. Reverse Bits
class Solution: # @param n, an integer # @return an integer def reverseBits(self, n): # n为0，或者n为负数，效率问题 if n == 0: return 0 binary = bin(n) if n < 0: binary = list(binary.lstrip(‘-0b‘)) else: binary = list(binary.lstrip(‘0b‘)) binary = list(reversed(binary)) if len(binary) < 32: binary.extend([‘0‘] * (32-len(binary))) res = int("".join(binary), 2) return -res if n < 0 else res
class Solution: # @param n, an integer # @return an integer def reverseBits(self, n): return int(‘0b‘ + ‘{:032b}‘.format(n)[::-1], 2)
class Solution: # @param n, an integer # @return an integer def reverseBits(self, n): m = 0 for i in xrange(32): m = (m << 1) | (n & 1) n = n >> 1 return m
Python LeetCode（七）
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