6.08

Author: princewen

easy/二进制/342_Power of Four.py

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+"""
+
+Given an integer (signed 32 bits), write a function to check whether it is a power of 4.
+
+Example:
+Given num = 16, return true. Given num = 5, return false.
+
+Follow up: Could you solve it without loops/recursion?
+
+
+
+"""
+
+class Solution(object):
+    def isPowerOfFour(self, num):
+        """
+        :type num: int
+        :rtype: bool
+        """
+        return num>0 and (num & num-1)==0 and (num-1)%3==0

easy/其他/290_Word Pattern.py

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+"""
+
+Given a pattern and a string str, find if str follows the same pattern.
+
+Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str.
+
+Examples:
+pattern = "abba", str = "dog cat cat dog" should return true.
+pattern = "abba", str = "dog cat cat fish" should return false.
+pattern = "aaaa", str = "dog cat cat dog" should return false.
+pattern = "abba", str = "dog dog dog dog" should return false.
+Notes:
+You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.
+
+"""
+
+"""my solution
+使用zip和set
+"""
+
+
+class Solution(object):
+    def wordPattern(self, pattern, str):
+        """
+        :type pattern: str
+        :type str: str
+        :rtype: bool
+        """
+
+        return len(pattern) == len(str.split(' ')) and len(set(zip(pattern, str.split(' ')))) == len(
+            set(pattern)) == len(set(str.split(' ')))

easy/字符串/344_Reverse String.py

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+"""
+
+Write a function that takes a string as input and returns the string reversed.
+
+Example:
+Given s = "hello", return "olleh".
+
+"""
+
+class Solution(object):
+    def reverseString(self, s):
+        """
+        :type s: str
+        :rtype: str
+        """
+        return s[::-1]

easy/字符串/__init__.py

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+"""
+
+Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.
+
+For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].
+
+Note:
+You must do this in-place without making a copy of the array.
+Minimize the total number of operations.
+
+
+"""
+
+"""my solution
+先把不为零的往前放，最后补0
+"""
+
+class Solution(object):
+    def moveZeroes(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: void Do not return anything, modify nums in-place instead.
+        """
+        index = 0
+        for num in nums:
+            if num != 0:
+                nums[index] = num
+                index += 1
+        for i in range(index,len(nums)):
+            nums[i] = 0

easy/数组/283_Move Zeroes.py

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+"""
+
+Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
+
+Example:
+Given nums = [-2, 0, 3, -5, 2, -1]
+
+sumRange(0, 2) -> 1
+sumRange(2, 5) -> -1
+sumRange(0, 5) -> -3
+Note:
+You may assume that the array does not change.
+There are many calls to sumRange function.
+
+"""
+"""这里用到的技巧就是数组存储前n项的累加和，这里要判断i是否为0"""
+
+class NumArray(object):
+    def __init__(self, nums):
+        """
+        :type nums: List[int]
+        """
+        self.dp = nums
+        for i in range(1, len(nums)):
+            self.dp[i] += self.dp[i - 1]
+
+    def sumRange(self, i, j):
+        """
+        :type i: int
+        :type j: int
+        :rtype: int
+        """
+        return self.dp[j] - (self.dp[i - 1] if i > 0 else 0)

easy/数组/303_Range Sum Query - Immutable.py

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+"""
+
+Given two arrays, write a function to compute their intersection.
+
+Example:
+Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].
+
+Note:
+Each element in the result must be unique.
+The result can be in any order.
+
+"""
+
+"""my solution"""
+
+
+class Solution(object):
+    def intersection(self, nums1, nums2):
+        """
+        :type nums1: List[int]
+        :type nums2: List[int]
+        :rtype: List[int]
+        """
+
+        nums1 = set(nums1)
+        return [x for x in set(nums2) if x in nums1]
+
+"""other solution
+直接用set的基本操作：
+>>> x = set("jihite")
+>>> y = set(['d', 'i', 'm', 'i', 't', 'e'])
+>>> x       #把字符串转化为set，去重了
+set(['i', 'h', 'j', 'e', 't'])
+>>> y
+set(['i', 'e', 'm', 'd', 't'])
+>>> x & y   #交
+set(['i', 'e', 't'])
+>>> x | y   #并
+set(['e', 'd', 'i', 'h', 'j', 'm', 't'])
+>>> x - y   #差
+set(['h', 'j'])
+>>> y - x
+set(['m', 'd'])
+>>> x ^ y   #对称差：x和y的交集减去并集
+set(['d', 'h', 'j', 'm'])
+
+"""
+
+def intersection(self, nums1, nums2):
+    """
+    :type nums1: List[int]
+    :type nums2: List[int]
+    :rtype: List[int]
+    """
+    nums1=set(nums1)
+    nums2=set(nums2)
+    return list(nums1&nums2)