Solution 7: 判断两链表是否相交
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Solution 7: 判断两链表是否相交
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問題描述
RT.
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解決思路
(1)?兩鏈表都是單向鏈表：判斷兩鏈表的末尾節點是否相同；
(2)?兩鏈表中一個有環，一個沒環：不可能相交；
(3)?兩鏈表都有環：slow-fast雙指針方法。
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程序
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public class ListIntersection {// two single listpublic boolean isIntersectionOfTwoSingleList(ListNode l1, ListNode l2) {if (l1 == null || l2 == null) {return false;}// whether the end of two list is sameListNode endOfList1 = getEndOfList(l1);ListNode endOfList2 = getEndOfList(l2);return endOfList1 == endOfList2;}private ListNode getEndOfList(ListNode head) {if (head == null) {return null;}ListNode node = head;while (node.next != null) {node = node.next;}return node;}// two list with cyclepublic boolean isIntersectionOfTwoListWithCycle(ListNode l1, ListNode l2) {if (l1 == null || l2 == null) {return false;}ListNode slow = l1, fast = l2;while (fast.next != null || fast != null || slow != null) {slow = slow.next;fast = fast.next.next;if (slow == fast) {return true;}}return false;} }?
Follow?up
求出兩個鏈表相交的第一個節點（如果存在的話）。
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(1)?兩條單鏈表
a. 求出兩條鏈表的長度及長度之差diff，然后設立兩個指針指向兩鏈表的頭結點，其中指向長鏈表頭結點的指針向前移動diff步；
b. 然后同時移動兩指針，直到所指節點相同（地址相同）為止，否則返回null。
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(2)?兩條鏈表有環
首先slow-fast，直到相遇為止，其中任意一個指針指回其頭結點，然后slow和fast指針同時移動，直到相遇，相遇的節點為第一個相交的節點。
（注意：可能有兩個相交的節點）
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程序
public class ListIntersection2 {// two single listpublic ListNode getFirstIntersectionNodeOfSingleList(ListNode l1,ListNode l2) {ListNode longNode = l1, shortNode = l2;int len1 = getLenOfList(l1);int len2 = getLenOfList(l2);if (len1 < len2) {longNode = l2;shortNode = l1;}int diff = Math.abs(len1 - len2);// long move diff stepswhile (diff > 0) {longNode = longNode.next;--diff;}while (longNode != null && shortNode != null) {if (longNode == shortNode) {return longNode;}longNode = longNode.next;shortNode = shortNode.next;}return null;}private int getLenOfList(ListNode head) {ListNode node = head;int len = 0;while (node != null) {++len;node = node.next;}return len;}// two list with cyclepublic ListNode getFirstIntersectionNodeOfCycleList(ListNode l1, ListNode l2) {if (l1 == null || l2 == null) {return null;}ListNode slow = l1, fast = l2;while (fast.next != null || fast != null || slow != null) {if (fast == slow) {break;}slow = slow.next;fast = fast.next.next;}if (fast == null || slow == null) {return null;}slow = l1;while (slow != fast) {slow = slow.next;fast = fast.next;}return slow;} }?
轉載于:https://www.cnblogs.com/harrygogo/p/4614277.html
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