兩個鏈表中的第一個公共頂點:
解法一:兩次遍歷即可
第一次遍歷找到兩個鏈表的長度,求出差值k,然後較長的;鏈表先走k步,之後兩個鏈表同時走,直到遇到第一個相同的結點爲止
解法二:輔助棧,先順序遍歷並將鏈表內容存儲到棧中,然後依次彈棧,直到遇到最後一個不同結點
public class _Q37<T> {
public ListNode<T> FindFirstCommonNode(ListNode<T> list1, ListNode<T> list2){
if(list1 == null || list2 == null) return null;
ListNode<T> node1 = list1;
ListNode<T> node2 = list2;
int len1 = 0;
int len2 = 0;
while(node1 != null) {node1 = node1.next; len1++;}
while(node2 != null) {node2 = node2.next; len2++;}
node1 = list1;
node2 = list2;
int k = 0;
if(len1 >= len2){
k = len1 - len2;
while(k > 0){ node1 = node1.next; --k;}
}else{
k = len2 - len1;
while(k > 0){ node2 = node2.next; --k;}
}
while (node1 != null && node2 != null && node1.value != node2.value) {
node1 = node1.next;
node2 = node2.next;
}
return node1; // node可能爲空,因爲可能輸入鏈表中並沒有相同結點
}
}
測試代碼:
public class _Q37Test extends TestCase {
_Q37<Integer> firstCommon = new _Q37<Integer>();
public void test(){
// 一旦第一個相同,後面都必須相同
int array1[] = {1, 2, 3, 4, 5, 6};
int array2[] = {6};
int array3[] = {7, 8, 9, 10};
ListNode<Integer> list1 = new ListNode<Integer>();
ListNode<Integer> list2 = new ListNode<Integer>();
ListNode<Integer> list3 = new ListNode<Integer>();
ListNode<Integer> p = list1;
for(int i=0; i<array1.length; i++){
ListNode<Integer> node = new ListNode<>();
node.value = array1[i];
node.next = p.next;
p.next = node;
p = p.next;
}
p = list2;
for(int i=0; i<array2.length; i++){
ListNode<Integer> node = new ListNode<>();
node.value = array2[i];
node.next = p.next;
p.next = node;
p = p.next;
}
p = list3;
for(int i=0; i<array2.length; i++){
ListNode<Integer> node = new ListNode<>();
node.value = array3[i];
node.next = p.next;
p.next = node;
p = p.next;
}
System.out.println(firstCommon.FindFirstCommonNode(list1.next, list2.next).value);
if(firstCommon.FindFirstCommonNode(list1.next, list3.next) != null){
System.out.println(firstCommon.FindFirstCommonNode(list1.next, list3.next).value);
}else{
System.out.println("null");
}
}
}