題目描述
輸入兩個鏈表,找出它們的第一個公共結點。
解題思路
看到這道題之後,第一反應是用利用兩個鏈表的長度差來做。如果兩個鏈表有公共節點,那兩個鏈表共用公共節點之後的部分。計算兩個鏈表的長度差diff
,讓較長的鏈表前進diff
後,這兩個鏈表同時同步向後移動,直至兩個鏈表的節點相等。
def FindFirstCommonNode(self, pHead1, pHead2):
len1 = self.getChainLen(pHead1)
len2 = self.getChainLen(pHead2)
if len2 > len1:
pHead1, pHead2 = pHead2, pHead1
diff = abs(len1-len2)
while diff > 0:
pHead1 = pHead1.next
diff -= 1
while pHead1 != pHead2:
pHead1 = pHead1.next
pHead2 = pHead2.next
return pHead1
def getChainLen(self, Head):
chainLen = 0
while Head:
chainLen += 1
Head = Head.next
return chainLen
之後在搜尋更優解的時候,看到了如下這段代碼,對其巧妙的思路折服!時間複雜度和上述代碼相同,都需要遍歷鏈表(1+公共節點之前的長度/總長度)
遍,但短的代碼長度,更精妙的解題思路,更少的變量,美不勝收。。
話不多說:
def FindFirstCommonNode(self, pHead1, pHead2):
p1, p2 = pHead1, pHead2
while p1 != p2:
p1 = p1.next if p1 != None else pHead2
p2 = p2.next if p2 != None else pHead1
return p1