3.寻找重复数字
public boolean findRepeatNum(int[] nums){
HashSet<Integer> set = new HashSet();
for (Integer i : nums){
if(set.contains(i)){
return true;
}else{
set.add(i);
}
}
return false;
}
4.查找二维数组中的值
public boolean findNumInTrix(int[][] nums,int num){
int i = 0,int j = nums[0].length;
while (i<=nums.length&&j>=0){
if(num < nums[i][j]){
j--;
}else if(num> nums[i][j]) {
i++
}else{
return true;
}
}
return false;
}
5.替换空格
public String replaceWhiteSpace(){
StringBuilder sb = new StringBuilder();
for (Character c:sb.toCharArray()){
if (c==' '){
sb.append("20%")
}
sb.append(c);
}
return sb.toString();
}
6.反向打印链表
方法一：辅助栈法
- 这种方法真的非常巧妙，后加入的节点后交换位置，完美的诠释了栈的作用；
Stack<Integer> stack = new Stack();
public void reverseListNode(ListNode head){
while(head!= null){
stack.push(head.val);
head = head.next;
}
}
public void print(){
while (stack.size()!=0){
System.out.println(stack.pop);
}
}
方法二：递归法
LinkedList<Integer> list = new LinkedList();
public void printListNode(ListNode head){
recur(head);
for(int i = 0;i<list.lenth;i++){
System.out.println(list.get(i));
}
}
//递归将数字装入list
public void recur(ListNode head){
if(head == null){
return;
}
recur(head.next);
list.add(head.val);
}
7.重建二叉树-是一个不断寻找左右子树的过程
- 制作中序遍历的下标集合
- 递归方法
- 边界判断
class Solution {
int[] preorder;
HashMap<Integer, Integer> dic = new HashMap<>();
public TreeNode buildTree(int[] preorder, int[] inorder) {
this.preorder = preorder;
for(int i = 0; i < inorder.length; i++)
dic.put(inorder[i], i);
return recur(0, 0, inorder.length - 1);
}
//root:先序遍历的索引；left：中序遍历左边界；right：中序遍历右边界
TreeNode recur(int root, int left, int right) {
if(left > right) return null; // 递归终止
TreeNode node = new TreeNode(preorder[root]); // 建立根节点
//中序遍历索引根索引
int i = dic.get(preorder[root]); // 划分根节点、左子树、右子树
node.left = recur(root + 1, left, i - 1); // 开启左子树递归
node.right = recur(root + i - left + 1, i + 1, right); // 开启右子树递归
return node; // 回溯返回根节点
}
}
8.两个栈实现队列
class soluation{
private Stack<Integer> stackIn;
private Stack<Integer> stackOUt;
public Queue(){
stackIn = new Stack<Integer>;
stackOut = new Stack<Integer>;
}
public void queueIn(Integer i){
stackIn.add(i);
}
public Integer queueOut(){
if(stackOut.size()!=0){
return stackOut.pop();
}
if(stackIn.size()==0){
return null;
}
while (stackIn.size!=0){
stackOut.push(stackIn.pop);
}
return stackOut.pop;
}
}
10.斐波那契数列
public Integer fib(Integer num){
int fib0=0;
int fib1=1;
int tmp = 0;
for(int i=0 ; i<num ; i++){
tmp = (fib0 + fib1)%1000000007;
fib0 = fib1;
fib1 = tmp;
}
return fib1;
}
11.旋转数组的最小数字-二分法
public Integer findMinNum(int[] nums){
int i =0,j=nums.length;
int m = (i+j)/2;
while (i<j){
if(nums[m]<nums[j]){
j = m;
}else if(nums[m]>[j]){
i==m+1;
}else{
j--;
}
}
retrun nums[i];
}
13.矩阵中的路径
public boolean exist(char[][] board, String word) {
char[] words = word.toCharArray();
for(int i = 0; i < board.length; i++) {
for(int j = 0; j < board[0].length; j++) {
if(dfs(board, words, i, j, 0)) return true;
}
}
return false;
}
boolean dfs(char[][] board, char[] word, int i, int j, int k) {
if(i >= board.length || i < 0 || j >= board[0].length || j < 0 || board[i][j] != word[k]) return false;
if(k == word.length - 1) return true;
board[i][j] = ';'boolean
= res dfs (,board, word+ i 1 ,, j+ k 1 )|| dfs (,board, word- i 1 ,, j+ k 1 )|| dfs
(,board, word, i+ j 1 ,+ k 1 )|| dfs (,board, word, i - j 1 ,+ k 1 );[
board]i[]j= [ word]k;return
; res}
public
14.剪绳子
核心思想：3最优，2次优，1最差
int cuttingRope (int) nif {
(<n3)return ; n=
x % n3;=
y / n3;if
(==x0). reutrn Mathpow(3,)y;if
(==x2)return . Mathpow(3,-y1)*2;if
(==x1)return . Mathpow()}
public
15.二进制中“1”的个数
class Solution // you need to treat n as an unsigned value {
public
int hammingWeight (int) nint {
= res 0 ;while
( !=n0 )+={
res & n1;1
n>>>=;}
return
; res}
}
public
16.数值的整数次方
核心思想：就是一个降幂的过程。x是底数，n是指数，ans是结果，x^n代表的是一个数值，快速幂说白了就是在确保数值不变的情况下，用二分法把数值从指数不断地往底数里面搬。核心是n=n/2, x=x^2这两句代码。但是由于指数可能是奇数，整数除二会丢失一位底数，所以要在ans里面记录奇数轮丢失的底数。
double myPow (double, xint ) nif {
(==x 0 )return 0 ;long
= b ; ndouble
= res 1.0 ;if
(<b 0 )= {
x 1 / ; x=
b - ;b}
while
(0b > )if {
((&b 1 )== 1 )*= res ; x*=
x ; x1
b >>= ;}
return
; res}
public
17.打印从1到n的最大n位数
int []printNumbers (int) nint {
= end ( int).Mathpow(10,) n- 1 ;int
[]= res new int []end;for
(int= i 0 ;< i ; end++ i)[
res]i= + i 1 ;return
; res}
public
18.删除链表的节点
deleteNode ListNode (,ListNode headint ) valif {
( .head==val ) valreturn . head;next=
ListNode pre,head=cur.head;nextwhile
(!=cur&&null . cur!=val)val={
pre ; cur=
cur . cur;next}
.
pre=next . cur;nextreturn
; head}
class
19.正则表达式匹配
Solution public {
boolean isMatch (,String s) String pint {
= m . slength()+ 1 ,= n . plength()+ 1 ;boolean
[][]= dp new boolean []m[]n;[
dp0][0]= true ;for
(int= j 2 ;< j ; n+= j 2 )[
dp0][]j= [ dp0][-j 2 ]&& . pcharAt(-j 1 )== '*' ;for
(int= i 1 ;< i ; m++ i)for {
(int= j 1 ;< j ; n++ j)[ {
dp]i[]j= . pcharAt(-j 1 )== '*' ? [
dp]i[-j 2 ]|| [ dp-i 1 ][]j&& ( .scharAt(-i 1 )== . pcharAt(-j 2 )|| . pcharAt(-j 2 )== '.' ): [
dp-i 1 ][-j 1 ]&& ( .pcharAt(-j 1 )== '.' || . scharAt(-i 1 )== . pcharAt(-j 1 ));}
}
return
[ dp-m 1 ][-n 1 ];}
}
class
20.表示数值的字符串（有限状态机）
Solution public {
boolean isNumber ()String s[ {
Map]= states new {
HashMap <(>)put {{ (' ',0 );put ('s',1 );put ('d',2 );put ('.',4 );} },// 0. new
HashMap <(>)put {{ ('d',2 );put ('.',4 );} },// 1. new
HashMap <(>)put {{ ('d',2 );put ('.',3 );put ('e',5 );put (' ',8 );} },// 2. new
HashMap <(>)put {{ ('d',3 );put ('e',5 );put (' ',8 );} },// 3. new
HashMap <(>)put {{ ('d',3 );} },// 4. new
HashMap <(>)put {{ ('s',6 );put ('d',7 );} },// 5. new
HashMap <(>)put {{ ('d',7 );} },// 6. new
HashMap <(>)put {{ ('d',7 );put (' ',8 );} },// 7. new
HashMap <(>)put {{ (' ',8 );} }// 8. }
;int
= p 0 ;char
; tfor
(char: c . stoCharArray())if {
('0'c >= && <= c '9' )= t 'd' ;else
if (==c '+' || == c '-' )= t 's' ;else
if (==c 'e' || == c 'E' )= t 'e' ;else
if (==c '.' || == c ' ' )= t ; celse
= t '?' ;if
(![states]p.containsKey()t)return false ;=
p ( int)[states]p.get()t;}
return
== p 2 || == p 3 || == p 7 || == p 8 ;}
}
class
21.调换数组顺序是奇数位于偶数之前
Solution public {
int []exchange (int[]) numsint {
= i 0 ,= j . nums-length 1 ,; tmpwhile
(<i ) jwhile {
(<i && j ( [nums]i& 1 )== 1 )++ i;while
(<i && j ( [nums]j& 1 )== 0 )-- j;=
tmp [ nums]i;[
nums]i= [ nums]j;[
nums]j= ; tmp}
return
; nums}
}
class
22.链表中倒数第k个值
Solution public {
getKthFromEnd ListNode (,ListNode headint ) k= {
ListNode former , head= latter ; headfor
(int= i 0 ;< i ; k++ i)=
former . former;nextwhile
(!=former ) null= {
former . former;next=
latter . latter;next}
return
; latter}
}
class
24.反转链表
方法一：迭代（双指针）
Solution public {
reverseList ListNode ()ListNode head= {
ListNode cur , head= pre ; nullwhile
(!=cur ) null= {
ListNode tmp . cur;next// 暂存后继节点 cur.next .
cur=next ; pre// 修改 next 引用指向 =
pre ; cur// pre 暂存 cur =
cur ; tmp// cur 访问下一节点 }
return
; pre}
}
class
方法二：递归
Solution public {
reverseList ListNode ()ListNode headreturn {
recur (,head) null;// 调用递归并返回 }
private
recur ListNode (,ListNode cur) ListNode preif {
( ==cur ) nullreturn ; pre// 终止条件 =
ListNode res recur (.cur,next) cur;// 递归后继节点 .
cur=next ; pre// 修改节点引用指向 return
; res// 返回反转链表的头节点 }
}
class
25.合并两个排序链表
Solution public {
mergeTwoLists ListNode (,ListNode l1) ListNode l2= {
ListNode dum new ListNode (0),= cur ; dumwhile
(!=l1 && null != l2 ) nullif {
(.l1<val . l2)val. {
cur=next ; l1=
l1 . l1;next}
else
. {
cur=next ; l2=
l2 . l2;next}
=
cur . cur;next}
.
cur=next != l1 ? null : l1 ; l2return
. dum;next}
}
class
26.树的子结构
Solution public {
boolean isSubStructure (,TreeNode A) TreeNode Breturn {
( !=A && null != B ) null&& ( recur(,A) B|| isSubStructure (.A,left) B|| isSubStructure (.A,right) B);}
boolean
recur (,TreeNode A) TreeNode Bif {
(==B ) nullreturn true ;if
(==A || null . A!=val . B)valreturn false ;return
recur (.A,left. B)left&& recur (.A,right. B)right;}
}
class
27.二叉树的镜像
方法一：递归
Solution public {
mirrorTree TreeNode ()TreeNode rootif {
(==root ) nullreturn ; null=
TreeNode tmp . root;left.
root=left mirrorTree (.root)right;.
root=right mirrorTree ()tmp;return
; root}
}
class
方法二：辅助栈
Solution public {
mirrorTree TreeNode ()TreeNode rootif {
(==root ) nullreturn ; null<
Stack=TreeNode> stack new Stack <(>)add {{ ()root;} };while
(!.stackisEmpty())= {
TreeNode node . stackpop();if
(.node!=left ) null. stackadd(.node)left;if
(.node!=right ) null. stackadd(.node)right;=
TreeNode tmp . node;left.
node=left . node;right.
node=right ; tmp}
return
; root}
}
class
28.对称的二叉树
Solution public {
boolean isSymmetric ()TreeNode rootreturn {
== root ? null true : recur (.root,left. root)right;}
boolean
recur (,TreeNode L) TreeNode Rif {
(==L && null == R ) nullreturn true ;if
(==L || null == R || null . L!=val . R)valreturn false ;return
recur (.L,left. R)right&& recur (.L,right. R)left;}
}
class
29.顺时针打印矩阵
Solution public {
int []spiralOrder (int[][]) matrixif {
(.matrix==length 0 )return new int [0];int
= l 0 ,= r [ matrix0].-length 1 ,= t 0 ,= b . matrix-length 1 ,= x 0 ;int
[]= res new int [(+r 1 )* ( +b 1 )];while
(true)for {
(int= i ; l<= i ; r++ i)[ res++x]= [ matrix]t[]i;// left to right. if
(++)t > bbreak ;for
(int= i ; t<= i ; b++ i)[ res++x]= [ matrix]i[]r;// top to bottom. if
(--l > )rbreak ;for
(int= i ; r; i >= l-- i)[ res++x]= [ matrix]b[]i;// right to left. if
(--t > )bbreak ;for
(int= i ; b; i >= t-- i)[ res++x]= [ matrix]i[]l;// bottom to top. if
(++)l > rbreak ;}
return
; res}
}
class
30.包含main函数的栈
MinStack < {
Stack,Integer> A; Bpublic
MinStack ()= {
A new Stack <(>);=
B new Stack <(>);}
public
void push (int) x. {
Aadd()x;if
(.Bempty()|| . Bpeek()) >= x.
Badd()x;}
public
void pop ()if {
(.Apop().equals(.Bpeek())).
Bpop();}
public
int top ()return {
. Apeek();}
public
int min ()return {
. Bpeek();}
}
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