92 lines
1.6 KiB
Go
92 lines
1.6 KiB
Go
package _6_linkedlist
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/*
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思路1：开一个栈存放链表前半段
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时间复杂度：O(N)
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空间复杂度：O(N)
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*/
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func isPalindrome1(l *LinkedList) bool {
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lLen := l.length
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if lLen == 0 {
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return false
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}
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if lLen == 1 {
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return true
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}
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s := make([]string, 0, lLen/2)
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cur := l.head
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for i := uint(1); i <= lLen; i++ {
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cur = cur.next
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if lLen%2 != 0 && i == (lLen/2+1) { //如果链表有奇数个节点，中间的直接忽略
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continue
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}
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if i <= lLen/2 { //前一半入栈
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s = append(s, cur.GetValue().(string))
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} else { //后一半与前一半进行对比
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if s[lLen-i] != cur.GetValue().(string) {
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return false
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}
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}
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}
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return true
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}
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/*
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思路2
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找到链表中间节点，将前半部分转置，再从中间向左右遍历对比
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时间复杂度：O(N)
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*/
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func isPalindrome2(l *LinkedList) bool {
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lLen := l.length
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if lLen == 0 {
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return false
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}
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if lLen == 1 {
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return true
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}
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var isPalindrome = true
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step := lLen / 2
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var pre *ListNode = nil
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cur := l.head.next
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next := l.head.next.next
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for i := uint(1); i <= step; i++ {
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tmp := cur.GetNext()
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cur.next = pre
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pre = cur
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cur = tmp
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next = cur.GetNext()
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}
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mid := cur
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var left, right *ListNode = pre, nil
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if lLen%2 != 0 { //
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right = mid.GetNext()
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} else {
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right = mid
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}
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for nil != left && nil != right {
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if left.GetValue().(string) != right.GetValue().(string) {
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isPalindrome = false
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break
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}
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left = left.GetNext()
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right = right.GetNext()
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}
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//复原链表
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cur = pre
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pre = mid
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for nil != cur {
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next = cur.GetNext()
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cur.next = pre
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pre = cur
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cur = next
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}
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l.head.next = pre
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return isPalindrome
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}
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