ana

Author: liuyaqiao

kthLargestinTwoSortedArray.py

@@ -0,0 +1,71 @@
+'''
+K-th Element of Two Sorted Arrays
+Given two sorted arrays of size m and n respectively, you are tasked with finding the element that would be at the k’th position of the final sorted array
+
+Input : Array 1 - 2 3 6 7 9
+        Array 2 - 1 4 8 10
+        k = 5
+Output : 6
+Explanation: The final sorted array would be -
+1, 2, 3, 4, 6, 7, 8, 9, 10
+The 5th element of this array is 6.
+Input : Array 1 - 100 112 256 349 770
+        Array 2 - 72 86 113 119 265 445 892
+        k = 7
+Output : 256
+Explanation: Final sorted array is -
+72, 86, 100, 112, 113, 119, 256, 265, 349, 445, 770, 892
+7th element of this array is 256.
+'''
+
+
+def kthLargest(a,b,k):
+    c = []
+    i = 0
+    j = 0
+    while i < len(a) or j < len(b):
+        if a[i] < b[j]:
+            c.append(b[j])
+            j += 1
+        else:
+            c.append(a[i])
+            i += 1
+    
+    if i < len(a):
+        c.append(a[i])
+        i += 1
+    if j < len(b):
+        c.append(b[j])
+        j += 1
+    print(c)
+
+
+def findkth(a,b,k):
+
+    if not a:
+        return b[k]
+    if not b:
+        return a[k]
+
+    la, lb = len(a)//2, len(b)//2
+    ma, mb = a[la], a[lb]
+
+    if la + lb < k:
+        #第k大的值在右边，下面判断要去掉哪一部分的值
+        if ma > mb:
+            #说明肯定不在b的前半部分中
+            return findkth(a, b[lb:], k - ib)
+        else:
+            return findkth(a[:ia + 1], k - ia])
+    else:
+        if ma > mb:
+            return findkth(a[:ia], b, k)
+        else:
+            return findkth(a. b[:ib], k)
+
+
+
+if __name__ == '__main__':
+    a = [2,3,6,7,9]
+    b = [1,4,8,10]
+    print(findkth(a,b,3))·

src/252.py

@@ -32,4 +32,11 @@ def canAttendMeetings(self, intervals):
 我们只关心开始时间和结束时间，让它们排序，使用容器中元素的数量来进行判断；
 通过遍历时间，根据两个时间的差值来改变容器状态，通过状态来进行条件判断；
 
+intervals time --> greedy
+
+排序之后对我们的结果没有影响，因为可以转化为人进入房间的问题； 我们不需要知道
+每一个人进去出去的时间，我们只需要在某一个时刻有人进入，有人出去，通过他们的关系
+去判断。
+
+
 '''

src/367.py

@@ -0,0 +1,21 @@
+class Solution:
+    def isPerfectSquare(self, num):
+        """
+        :type num: int
+        :rtype: bool
+        """
+        if num < 2:
+            return True
+        
+        start = 1
+        end = num
+        
+        while start + 1 < end:
+            mid = start + (end - start) // 2
+            if mid * mid == num:
+                return True
+            elif mid * mid < num:
+                start = mid
+            else:
+                end = mid
+        return False

src/56.py

@@ -0,0 +1,25 @@
+vsclass Solution:
+    def mySqrt(self, x):
+        """
+        :type x: int
+        :rtype: int
+        """
+        if x < 2:
+            return x
+        
+        start = 1
+        end = x
+        while start + 1 < end:
+            mid = start + (end - start) // 2
+            if mid * mid == x:
+                return mid
+            elif mid * mid < x:
+                start = mid
+            else:
+                end = mid
+        
+        if start * start < x:
+            return start
+        
+        
+        

src/69.py

@@ -0,0 +1,25 @@
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def deleteDuplicates(self, head: 'ListNode') -> 'ListNode':
+        if not head:
+            return None
+        
+        preHead = ListNode(-1)
+        preHead.next = head
+        pre = preHead
+        tmp = head
+        while pre.next != None and pre.next.next != None:
+            if pre.next.val == pre.next.next.val:
+                sameNums = pre.next.val
+                while pre.next != None and pre.next.val == sameNums:
+                    pre.next = pre.next.next
+            else:
+                pre = pre.next
+                
+        return preHead.next
+