2019-09-08

0001.两数之和/0001-两数之和.py

@@ -0,0 +1,15 @@
+class Solution(object):
+    def twoSum(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: List[int]
+        """
+        hashmap = {}
+        for i, x in enumerate(nums):
+            if hashmap.has_key(target - x):
+                return [hashmap[target - x], i]
+            else:
+                hashmap[x] = i
+
+        return []

0002.两数相加/0002-两数相加.py

@@ -0,0 +1,39 @@
+# Definition for singly-linked list.
+# class ListNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution(object):
+    def addTwoNumbers(self, l1, l2):
+        """
+        :type l1: ListNode
+        :type l2: ListNode
+        :rtype: ListNode
+        """
+        if self.getLength(l1) < self.getLength(l2):
+            l1, l2 = l2, l1
+        head = l1
+        while(l2):
+            l1.val += l2.val
+            l1 = l1.next
+            l2 = l2.next
+
+        p = head
+        while(p):
+            if p.val > 9:
+                p.val -= 10
+                if p.next:
+                    p.next.val += 1
+                else:
+                    p.next = ListNode(1)
+            p = p.next   
+        return head
+
+
+    def getLength(self, l):
+        tmp = 0
+        while(l):
+            tmp += 1
+            l = l.next
+        return tmp

0003.无重复字符的最长子串/0003-无重复字符的最长子串.py

@@ -0,0 +1,16 @@
+class Solution(object):
+    def lengthOfLongestSubstring(self, s):
+        """
+        :type s: str
+        :rtype: int
+        """
+        record = {}
+        start, res = 0, 0
+        for end in range(len(s)):
+            if s[end] in record:
+                start = max(start, record[s[end]] + 1)
+
+            record[s[end]] = end
+            res = max(res, end - start + 1)
+
+        return res

0004.寻找两个有序数组的中位数/0004-寻找两个有序数组的中位数.py

@@ -1,46 +1,22 @@
 from heapq import *
-class MedianFinder(object):
-# 维护两个堆，一个大顶堆，一个小顶堆，小顶堆里的数比大顶堆里的数都要大， 
-# 如果有两个潜在的中位数（两个堆size相同），数据流的中位数就是两个堆顶之和除以2
-# 如果只有一个中位数，就看size更小的那个堆的堆顶
-# 如果新进来的数比小顶堆的数要小，就把它插入大顶堆
-# 如果新进来的数比小顶堆的数要大，就把它插入小顶堆
-# 调整两个堆，使得size 差最大为1
+class DoubleHeap(object):
     def __init__(self):
-        """
-        initialize your data structure here.
-        """
-        self.max_h = list()
-        self.min_h = list()
-        heapify(self.max_h)
-        heapify(self.min_h)
+        self.maxh = []
+        self.minh = []
+        heapify(self.maxh)
+        heapify(self.minh)
 
-
-    def addNum(self, num):
-        """
-        :type num: int
-        :rtype: None
-        """
-        heappush(self.min_h, num)
-        heappush(self.max_h, -heappop(self.min_h))
-        if len(self.max_h) > len(self.min_h):
-            heappush(self.min_h, -heappop(self.max_h))
-
-    def findMedian(self):
-        """
-        :rtype: float
-        """
-        max_len = len(self.max_h)
-        min_len = len(self.min_h)
-        if max_len == min_len: #有两个候选中位数
-            return (self.min_h[0] + -self.max_h[0]) / 2.
-        else:#小顶堆的size 一定 >= 大顶堆的size，所以答案就是小顶堆的堆顶
-            return self.min_h[0] / 1.
+    def insert(self, val):
+        heappush(self.minh, val)
+        heappush(self.maxh, -heappop(self.minh))
+        if len(self.maxh) > len(self.minh):
+            heappush(self.minh, -heappop(self.maxh))
 
-# Your MedianFinder object will be instantiated and called as such:
-# obj = MedianFinder()
-# obj.addNum(num)
-# param_2 = obj.findMedian()
+    def findMedian(self):
+        if len(self.maxh) == len(self.minh):
+            return (self.minh[0] - self.maxh[0]) / 2.0
+        return self.minh[0]/1.0
+
 
 class Solution(object):
     def findMedianSortedArrays(self, nums1, nums2):
@@ -49,9 +25,11 @@ def findMedianSortedArrays(self, nums1, nums2):
         :type nums2: List[int]
         :rtype: float
         """
-        mf = MedianFinder()
+        dh = DoubleHeap()
         for num in nums1:
-            mf.addNum(num)
+            dh.insert(num)
+
         for num in nums2:
-            mf.addNum(num)
-        return mf.findMedian()
+            dh.insert(num)
+
+        return dh.findMedian()

0005.最长回文子串/0005-最长回文子串.py

@@ -4,24 +4,30 @@ def longestPalindrome(self, s):
         :type s: str
         :rtype: str
         """
-        max_l = 0
-        res = ""
-        for i in range(0, len(s)):
-            #以s[i] 为中心向左右扩散
+        left, right = 0, 0
+        res, string = 0, ""
+        for i in range(len(s)):
             left, right = i, i
             while(left >= 0 and right < len(s) and s[left] == s[right]):
-                if max_l < right - left + 1:
-                    max_l = right - left + 1
-                    res = s[left:right + 1]
                 left -= 1
                 right += 1
-
-            #以s[i],s[i+1]为中心向左右扩散
-            left, right = i, i + 1
+            left += 1
+            right -= 1
+
+            if right - left + 1 > res:
+                res = right - left + 1
+                string = s[left:right + 1]
+
+        for i in range(1, len(s)):
+            left, right = i - 1, i
             while(left >= 0 and right < len(s) and s[left] == s[right]):
-                if max_l < right - left + 1:
-                    max_l = right - left + 1
-                    res = s[left:right + 1]
                 left -= 1
-                right += 1            
-        return res
+                right += 1
+            left += 1
+            right -= 1
+
+            if right - left + 1 > res:
+                res = right - left + 1
+                string = s[left:right + 1]      
+        return string
+

0006.Z字形变换/0006-Z字形变换.py

@@ -1,35 +1,40 @@
 class Solution(object):
-    def convert(self, s, n):
+    def convert(self, s, numRows):
         """
         :type s: str
         :type numRows: int
         :rtype: str
         """
-        #ÕÒ¹æÂÉ
-        if n <= 1:
+        #第一行和最后一行都是相差 2 * (n - 1)
+        #对于直角在上面的直角三角形， 相差 2 * (n - 1 - i)
+        #对于直角在下面的直角三角形， 相差 2 * i
+        if not s or numRows == 1:
             return s
-        l = len(s)
         res = ""
-        for i in range(n):
-            tmp, index = "", i
-            if i in [0, n - 1]:
-                while(index < l):
-
-                    tmp += s[index]
-                    index += 2 * (n - 1)
+        for idx in range(numRows):
+            if idx < len(s):
+                res += s[idx]
+
+            if idx in [0, numRows - 1]:
+                tmp = idx + 2 *(numRows - 1)
+                while tmp < len(s):
+                    res += s[tmp]
+                    tmp += 2 *(numRows - 1)
             else:
-                state = "down"
-                while(index < l):
-                    tmp += s[index]
-                    if state == "down":
-                        state = "up"
-                        index += 2 * (n - 1 - i)
-                    else:
-                        state = "down"
-                        index += 2 * i
-            res += tmp
-
+                tmp = idx + 2 * (numRows - 1 - idx)
+                tri = "down"
+                while tmp < len(s):
+                    res += s[tmp]
+                    if tri == "up":
+                        tmp += 2 * (numRows - 1 - idx)
+                        tri = "down"
+                    else:       
+                        tmp += 2 * idx
+                        tri = "up"
+                    
         return res
 
-
+
+
+
 

0007.整数反转/0007-整数反转.py

@@ -4,19 +4,16 @@ def reverse(self, x):
         :type x: int
         :rtype: int
         """
-
-        flag = 0
+        INT_MIN = -2 **31
+        INT_MAX = 2 ** 31 - 1
+        op = 1
         if x < 0:
-            flag = 1
-        if flag:
+            op = -1
             s = str(x)[1:]
-            s = s[::-1]
-            x = -1 *int(s)            
         else:
             s = str(x)
-            s = s[::-1]
-            x = int(s)
 
-        if x < -1 * 2 **31 or x > 2** 31 -1:
-            return 0    
-        return x
+        res = op * int(s[::-1])
+        return res if INT_MIN <= res <= INT_MAX else 0
+
+

0008.字符串转换整数(atoi)/0008-字符串转换整数(atoi).py

@@ -5,32 +5,34 @@ def myAtoi(self, s):
         :rtype: int
         """
         s = s.strip(" ")
-        if not len(s): #排除空
-            return 0
-        if s[0] not in ["+", "-"] and not s[0].isdigit(): #排除第一个非空字符不是数字
+        # print s
+        if not s or (s[0] not in ["+", "-"] and not s[0].isdigit()):
             return 0
+
         op = 1
-        res = ""
+        tmp = ""
         for i, char in enumerate(s):
-            if i == 0 :
+            if i == 0:
                 if char == "-":
                     op = -1
                     continue
                 elif char == "+":
-                    continue
-            if char == " " or not char.isdigit():
+                    pass
+                    continue       
+            if char.isdigit():
+                tmp += char
+            else:
                 break
-            res += char
-        # print res, op
-        if len(res) > 0:
-            res = op * int(res)
+        # print tmp
+        if tmp:
+            res = op * int(tmp)
         else:
-            return 0
-        INT_MIN = -2 **31
+            res = 0
         INT_MAX = 2 **31 - 1
+        INT_MIN = -2 **31
         if res > INT_MAX:
             return INT_MAX
         elif res < INT_MIN:
             return INT_MIN
-        return res
-
+        else:
+            return res

0009.回文数/0009-回文数.py

@@ -1,11 +1,17 @@
 class Solution(object):
     def isPalindrome(self, x):
+        """
+        :type x: int
+        :rtype: bool
+        """
+        #2019.6.1
+        xx = x
+        if x < 0:
+            return False
 
-        s = str(x)
-        if len(s) <= 1:
-            return True
-        print len(s) / 2
-        for count in range(0, len(s)/2):
-            if s[count] != s[len(s)-1  - count]:
-                return False
-        return True
+        reverse = 0
+        while x > 0:
+            x, tmp = divmod(x, 10)
+            reverse = reverse * 10 + tmp
+
+        return reverse == xx