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Commitf9ae7f6

Browse files
committed
leetcode
1 parent2089348 commitf9ae7f6

12 files changed

+215
-193
lines changed

‎src/com/imood/msjava/HeapSort1.java

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‎src/com/imood/msjava/HeapSortMin.java

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‎src/com/imood/msjava/Test.java

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@@ -43,17 +43,5 @@ class ListNode {
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ListNode(intx) {val =x; }
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}
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publicstaticvoidmain(String[]args) {
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Stringabc =newString("abc");
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Stringabc1 =newString("abc");
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System.out.println(abc==abc1);
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System.out.println(abc.equals(abc1));
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}
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}

‎src/com/imood/msjava/TryTest.java

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packagecom.imood.msjava.leetcode;
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/**
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* @description:判断链表是否存在环
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* @author: 微信公众号：码上Java
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* @createDate: 2020/7/29/0029
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*/
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publicclassHasCycle_141 {
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/**
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* 使用双指针，一个指针每次移动一个节点，一个指针每次移动两个节点，如果存在环，那么这两个指针一定会相遇。
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* @param head
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* @return
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*/
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publicbooleanhasCycle(ListNodehead) {
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if(head==null){
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returnfalse;
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}
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ListNodelow=head;
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ListNodehigh=head.next;
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while (low!=null &&high!=null &&high.next!=null){
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if(low==high){
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returntrue;
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}
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low=low.next;
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high=high.next;
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}
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returnfalse;
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}
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}

‎src/com/imood/msjava/leetcode/MergeTwoLists.java

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@@ -22,11 +22,9 @@ public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
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if(l1==null) {
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returnl1;
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}
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if(l2==null) {
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returnl2;
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}
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if(l1.val<l2.val){
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l1.next=mergeTwoLists(l1.next,l2);
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returnl1;
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packagecom.imood.msjava.leetcode;
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/**
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* @description:归并两个有序的链表
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* @author: 微信公众号：码上Java
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* @createDate: 2020/7/29/0029
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*/
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publicclassMergeTwoLists_21 {
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/**
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* 归并两个有序的链表
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* @param l1
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* @param l2
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* @return
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*/
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publicListNodemergeTwoLists(ListNodel1,ListNodel2) {
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if (l1 ==null) {
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returnl2;
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}
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if (l2 ==null){
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returnl1;
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}
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if (l1.val <l2.val) {
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l1.next =mergeTwoLists(l1.next,l2);
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returnl1;
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}else {
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l2.next =mergeTwoLists(l1,l2.next);
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returnl2;
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}
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}
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}
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packagecom.imood.msjava.leetcode;
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/**
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* @description:
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* @author: 微信公众号：码上Java
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* @createDate: 2020/7/29/0029
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*/
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publicclassRemoveNthFromEnd_19 {
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/**
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* 删除链表的倒数第 n 个节点 双指针
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* @param head
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* @param n
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* @return
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*/
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publicListNoderemoveNthFromEnd(ListNodehead,intn) {
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ListNodefast =head;
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// 快指针先走n步
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while (n-- >0) {
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fast =fast.next;
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}
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if (fast ==null){
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returnhead.next;
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}
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ListNodeslow =head;
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// 快慢指针一起走
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while (fast.next !=null) {
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fast =fast.next;
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slow =slow.next;
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}
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//当fast.next == null 的时候，就到了
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slow.next =slow.next.next;
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returnhead;
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}
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}
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packagecom.imood.msjava.leetcode;
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/**
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* @description: 链表反转
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* @author: 微信公众号：码上Java
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* @createDate: 2020/7/29/0029
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*/
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publicclassReverseList_206 {
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/**
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* 递归解决 —— 链表反转
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* @param head
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* @return
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*/
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publicListNodereverseList(ListNodehead) {
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if (head ==null ||head.next ==null) {
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returnhead;
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}
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ListNodenext =head.next;
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ListNodenewHead =reverseList(next);
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next.next =head;
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head.next =null;
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returnnewHead;
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}
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}
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packagecom.imood.msjava.leetcode;
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/**
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* @description: 167. 两数之和 II - 输入有序数组
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* @author: 微信公众号：码上Java
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* @createDate: 2020/7/29/0029
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*/
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publicclassTwoSum_167 {
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/**
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* 题目描述：在有序数组中找出两个数，使它们的和为 target。
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* 思路：
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* 使用双指针，一个指针指向值较小的元素，一个指针指向值较大的元素。指向较小元素的指针从头向尾遍历，指向较大元素的指针从尾向头遍历。
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* 如果两个指针指向元素的和 sum == target，那么得到要求的结果；
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* 如果 sum > target，移动较大的元素，使 sum 变小一些；
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* 如果 sum < target，移动较小的元素，使 sum 变大一些。
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* 数组中的元素最多遍历一次，时间复杂度为 O(N)。只使用了两个额外变量，空间复杂度为 O(1)。
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*
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*
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* @param numbers
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* @param target
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* @return
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*/
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publicint[]twoSum(int[]numbers,inttarget) {
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if(numbers==null){
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returnnull;
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}
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intlow=0;
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inthigh=numbers.length-1;
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while (low<high){
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intsum=numbers[low]+numbers[high];
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if(sum==target){
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//返回数组下标+1
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returnnewint[]{low+1,high+1};
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}elseif(sum<target){
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low++;
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}else {
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high--;
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}
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}
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returnnull;
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}
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}

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