剑指Offer之Java算法习题精讲链表与字符串及数组

剑指Offer之Java算法习题精讲链表与字符串及数组

题目一

解法

/**

* Definition for singly-linked list.

* public class ListNode {

* int val;

* ListNode next;

* ListNode() {}

* ListNode(int val) { this.val = val; }

* ListNode(int val, ListNode next) { this.val = val; this.next = next; }

* }

*/

class Solution {

public ListNode removeElements(ListNode headhttp://, int val) {

ListNode q = new ListNode(-1);

ListNode a = q;

q.next = head;

while(q.next!=null){

if(q.next.val==val){

q.next = q.next.next;

}else{

q = q.next;

}

}

return a.next;

}

}

题目二

解法

class Solution {

public boolean isIsomorphic(String s, String t) {

if(s.length()!=t.length()){

return false;

}

for(int i =0;i

if(s.indexOf(s.charAt(i))!=t.indexOf(t.charAt(i))){

return false;

}

}

return true;

}

}

题目三

解法

/**

* Definition for singly-linked list.

* public class ListNode {

* int val;

* ListNode next;

* ListNode(int x) { val = x; }

* }

*/

class Solution {

public void deleteNode(ListNode node) {

node.val = node.next.val;

node.next = node.next.next;

}

}

题目四

解法

class Solution {

public boolean isAnagram(String s, String t) {

if(s.length()!=t.length()) return false;

ArrayList list1 = new ArrayList<>();

ArrayList list2 = new ArrayList<>();

for (int i = 0; i < s.length(); i++) {

list1.add(s.charAt(i));

}

for (int i = 0; i < t.length(); i++) {

list2.add(t.charAt(i));

}

Collections.sort(list1);

Collections.sort(list2);

for (int i = 0; i < list1.size(); i++) {

if (list1.get(i)!=list2.get(i)) return false;

}

return true;

}

}

//要熟悉API

class Solution {

public boolean isAnagram(String s, String t) {

if (s.length() != t.length()) {

return false;

}

char[] str1 = s.toCharArray();

char[] str2 = t.toCharArray();

Arrays.sort(str1);

Arrays.sort(str2);

return Arrays.equals(str1, str2);

}

}

题目五

解法

class Solution {

public int missingNumber(int[] nums) {

Arrays.sort(nums);

for (int i = 0; i < nums.length; i++) {

if(i!=nums[i]){

return i;

}

}

return nums.length;

}

}

题目六

解法

class Solution {

public boolean wordPattern(String pattern, String s) {

String[] split = s.split(" ");

if(split.length!=pattern.length()) return false;http://

int q = 0;

for(int i =0;i

l: for(int w = 0;w

if(split[w].equals(split[i])){

q = w;

break l;

}

}

if(pattern.indexOf(pattern.charAt(i))!=q){

return false;

}

}

return true;

}

}

if(s.indexOf(s.charAt(i))!=t.indexOf(t.charAt(i))){

return false;

}

}

return true;

}

}

题目三

解法

/**

* Definition for singly-linked list.

* public class ListNode {

* int val;

* ListNode next;

* ListNode(int x) { val = x; }

* }

*/

class Solution {

public void deleteNode(ListNode node) {

node.val = node.next.val;

node.next = node.next.next;

}

}

题目四

解法

class Solution {

public boolean isAnagram(String s, String t) {

if(s.length()!=t.length()) return false;

ArrayList list1 = new ArrayList<>();

ArrayList list2 = new ArrayList<>();

for (int i = 0; i < s.length(); i++) {

list1.add(s.charAt(i));

}

for (int i = 0; i < t.length(); i++) {

list2.add(t.charAt(i));

}

Collections.sort(list1);

Collections.sort(list2);

for (int i = 0; i < list1.size(); i++) {

if (list1.get(i)!=list2.get(i)) return false;

}

return true;

}

}

//要熟悉API

class Solution {

public boolean isAnagram(String s, String t) {

if (s.length() != t.length()) {

return false;

}

char[] str1 = s.toCharArray();

char[] str2 = t.toCharArray();

Arrays.sort(str1);

Arrays.sort(str2);

return Arrays.equals(str1, str2);

}

}

题目五

解法

class Solution {

public int missingNumber(int[] nums) {

Arrays.sort(nums);

for (int i = 0; i < nums.length; i++) {

if(i!=nums[i]){

return i;

}

}

return nums.length;

}

}

题目六

解法

class Solution {

public boolean wordPattern(String pattern, String s) {

String[] split = s.split(" ");

if(split.length!=pattern.length()) return false;http://

int q = 0;

for(int i =0;i

l: for(int w = 0;w

if(split[w].equals(split[i])){

q = w;

break l;

}

}

if(pattern.indexOf(pattern.charAt(i))!=q){

return false;

}

}

return true;

}

}

l: for(int w = 0;w

if(split[w].equals(split[i])){

q = w;

break l;

}

}

if(pattern.indexOf(pattern.charAt(i))!=q){

return false;

}

}

return true;

}

}

if(split[w].equals(split[i])){

q = w;

break l;

}

}

if(pattern.indexOf(pattern.charAt(i))!=q){

return false;

}

}

return true;

}

}

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