try

Author: OnlyChristmas

Python/detect-capital.py

@@ -0,0 +1,41 @@
+'''
+Given a word, you need to judge whether the usage of capitals in it is right or not.
+
+We define the usage of capitals in a word to be right when one of the following cases holds:
+
+All letters in this word are capitals, like "USA".
+All letters in this word are not capitals, like "leetcode".
+Only the first letter in this word is capital if it has more than one letter, like "Google".
+Otherwise, we define that this word doesn't use capitals in a right way.
+
+Example 1:
+Input: "USA"
+Output: True
+
+Example 2:
+Input: "FlaG"
+Output: False
+
+Note: The input will be a non-empty word consisting of uppercase and lowercase latin letters.
+'''
+
+class Solution:
+    def detectCapitalUse(self, word):
+        """
+        :type word: str
+        :rtype: bool
+        """
+
+        # Approach #1
+        # Time:  O(1)
+        # Space: O(1)
+        #
+        # if word.lower() == word or word.capitalize() == word or word == word.upper():
+            # return True
+        # return False
+
+        # Approach #2 直接在return 中判断，进一步简化语句                     注意特殊条件''
+        # Time:  O(1)
+        # Space: O(1)
+        #
+        return word.islower() or  word.isupper() or word.istitle() or word == ''

Python/implement-strstr.py

@@ -0,0 +1,71 @@
+'''
+Implement strStr().
+
+Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
+
+Example 1:
+Input: haystack = "hello", needle = "ll"
+Output: 2
+
+Example 2:
+Input: haystack = "aaaaa", needle = "bba"
+Output: -1
+Clarification:
+
+What should we return when needle is an empty string? This is a great question to ask during an interview.
+
+For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf().
+'''
+
+class Solution:
+    def strStr(self, haystack, needle):
+        """
+        :type haystack: str
+        :type needle: str
+        :rtype: int
+        """
+
+        # Approach #1 以主串为主干，进行一次遍历。边界条件比较多，不容易一次考虑周全。
+        # Time:  O()
+        # Space: O()
+        #
+#         substr_len =  len(needle)
+#         str_len = len(haystack)
+
+#         if substr_len  == 0: return 0
+#         if substr_len <= str_len:               # 考虑子串比主串更长的情况
+#             for i in range(str_len):            #  主字符串遍历一次
+#                 if haystack[i] == needle[0] and str_len - i >= substr_len:
+#                     for j in range(substr_len):
+#                         if  i+j >= str_len  or haystack[i+j] != needle[j] :          # 小心主串越界
+#                             break
+#                         elif haystack[i+j] == needle[j] and j == substr_len - 1:
+#                             return i
+#                 elif str_len - i < substr_len or i == str_len-1:      # 主串过短，无法完成匹配，提前终止。  主串匹配到了最后一个字符也无法匹配的特殊情况 "mississippi"  "a"
+#                     return -1
+#         else:
+            # return -1
+
+
+        # Approach #2 是不是有点投机的嫌疑？ 直接用字符串的内建函数s.find()
+        # Time:  O(1)
+        # Space: O(1)
+        #
+        # if(len(needle)==0):
+        #     return 0
+        # else:
+        #     return haystack.find(needle)
+
+
+        # Approach #3
+        # Time:  O(n)
+        # Space: O(1)
+        #
+        for i in range(len(haystack)-len(needle)+1):   # 有效减少比较的次数
+            if haystack[i:i+len(needle)]==needle:       #　字符串切片操作可以直接比较一个子串，不用每个字符都比较
+                return i                              # i 为 0 的时候， haystack[0:0] 表示为 0
+        return -1
+
+
+
+

Python/keyboard-row.py

@@ -0,0 +1,73 @@
+'''
+Given a List of words, return the words that can be typed using letters of alphabet on only one row's of American keyboard like the image below.
+
+
+American keyboard
+
+
+Example 1:
+Input: ["Hello", "Alaska", "Dad", "Peace"]
+Output: ["Alaska", "Dad"]
+
+Note:
+You may use one character in the keyboard more than once.
+You may assume the input string will only contain letters of alphabet.
+
+
+'''
+
+class Solution:
+    def findWords(self, words):
+        """
+        :type words: List[str]
+        :rtype: List[str]
+        """
+
+
+        #　method once  初始化太复杂了
+        # Time:  O(n^2)
+        # Space: O(n)
+        #
+#         res = []
+#         ref = [['q','w','e','r','t','y','u','i','o','p'],
+#               ['a','s','d','f','g','h','j','k','l'],
+#                ['z','x','c','v','b','n','m']]
+#         for i in words:
+#             ss = "".join(set(i.lower()))     # 要处理大小写问题
+#             for j in ref:
+#                 for k in ss:
+#                     if  k not in j:
+#                         break
+#                     elif k == ss[-1]:
+#                         res.append(i)
+#         return res
+
+
+        # method twice  用str的内置函数str.issubset()更加简洁
+        # Time:  O(n)
+        # Space: O(n)
+        #
+        # line1, line2, line3 = set('qwertyuiop'), set('asdfghjkl'), set('zxcvbnm')
+        # res = []
+        # for word in words:
+        #     ss = set(word.lower())
+        #     if ss.issubset(line1) or ss.issubset(line2) or ss.issubset(line3):
+        #         res.append(word)
+        # return res
+
+
+
+        # method third  本质上和方法二是一样的，只是用all()函数替换了issubset()函数
+        # Time:  O(n)
+        # Space: O(n)
+        #
+        line1, line2, line3 = set('qwertyuiop'), set('asdfghjkl'), set('zxcvbnm')
+        res = []
+        for word in words:
+            smallword = word.lower()
+            if all([x in line1 for x in smallword]) or all([x in line2 for x in smallword]) or all([x in line3 for x in smallword]):
+                res.append(word)
+        return res
+
+
+

Python/next-greater-element-i.py

@@ -0,0 +1,87 @@
+'''
+You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1's elements in the corresponding places of nums2.
+
+The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.
+
+Example 1:
+Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
+Output: [-1,3,-1]
+Explanation:
+    For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
+    For number 1 in the first array, the next greater number for it in the second array is 3.
+    For number 2 in the first array, there is no next greater number for it in the second array, so output -1.
+
+Example 2:
+Input: nums1 = [2,4], nums2 = [1,2,3,4].
+Output: [3,-1]
+Explanation:
+    For number 2 in the first array, the next greater number for it in the second array is 3.
+    For number 4 in the first array, there is no next greater number for it in the second array, so output -1.
+
+Note:
+All elements in nums1 and nums2 are unique.
+The length of both nums1 and nums2 would not exceed 1000.
+
+'''
+
+
+
+
+
+class Solution:
+    def nextGreaterElement(self, nums1, nums2):
+        """
+        :type nums1: List[int]
+        :type nums2: List[int]
+        :rtype: List[int]
+        """
+        # method once 双重循环效率较低
+        # Time:  O(nlogn)
+        # Space: O(n)
+        #
+        # res = []
+        # length = len(nums2)
+        # for i in nums1:
+        #     if (nums2.index(i)+1) < length:
+        #         for j in range(nums2.index(i)+1,length):
+        #             if nums2[j] > i:
+        #                 res.append(nums2[j])
+        #                 break
+        #             elif j == length-1:
+        #                 res.append(-1)
+        #     else:
+        #         res.append(-1)
+        # return res
+
+        # method twice  利用字典提升效率
+        # Time:  O(nlogn)
+        # Space: O(n)
+        #
+        # if nums2 != []:
+        #     dic  = {nums2[-1]:-1}
+        #     for index in range(len(nums2)) :
+        #         for j in range(index+1,len(nums2)):
+        #             if nums2[index] < nums2[j]:
+        #                 dic[ nums2[index]] = nums2[j]
+        #                 break
+        #             elif j == len(nums2)-1:
+        #                 dic[nums2[index]] = -1
+        #     return [dic.get(i,-1) for i in nums1]
+        # return []
+
+
+
+        # method third   利用栈结构提升效率
+        # Time:  O(n)
+        # Space: O(n)
+        #
+        dic , stack = {} ,[]
+        for number in nums2:
+            while stack and stack[-1] < number:
+                dic[stack.pop()] = number
+            stack.append(number)
+        return [dic.get(i , -1) for i in nums1]
+
+
+
+

Python/reverse-string-ii.py

@@ -0,0 +1,28 @@
+'''
+Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original.
+
+Example:
+Input: s = "abcdefg", k = 2
+Output: "bacdfeg"
+
+Restrictions:
+The string consists of lower English letters only.
+Length of the given string and k will in the range [1, 10000]
+
+'''
+
+class Solution:
+    def reverseStr(self, s, k):
+        """
+        :type s: str
+        :type k: int
+        :rtype: str
+        """
+        # Approach #1 字符串的切片和反序操作不能同时进行
+        # Time:  O(n)
+        # Space: O(1)
+
+        res = ''
+        for i in range(0,len(s),2*k):
+            res = res + s[i:i+k][::-1] + s[i+k:i+2*k]
+        return res