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283. 移动零

https://leetcode.cn/problems/move-zeroes/

class Solution {//方法一：简单覆盖//非零的从下标0添加，数组剩下位置都填0// public void moveZeroes(int[] nums) {//     int len=nums.length;//     int k=0;//     for(int i=0;i<len;i++){//         if(nums[i]!=0){//             nums[k++]=nums[i];//         }//     }//     while(k<len){//         nums[k++]=0;//     }// }//方法二：双指针//解释一下：如果数组没有0，那么快慢指针始终指向同一个位置，每个位置自己和自己交换；//如果数组有0，快指针先走一步，此时慢指针对应的就是0，所以要交换。public void moveZeroes(int[] nums) {int len=nums.length;int left=0;int right=0;while(right<len){if(nums[right]!=0){int temp = nums[left];nums[left] = nums[right];nums[right] = temp;left++;}right++;}}
}

11. 盛最多水的容器

https://leetcode.cn/problems/container-with-most-water/

class Solution {public int maxArea(int[] height) {//若向内 移动短板 ，水槽可能变大，因此下个水槽的面积 可能增大 。//若向内 移动长板 ，下个水槽的面积 可能会变小（或者不变） int left=0;int right=height.length-1;int res=0;while(left<right){int area= Math.min(height[left],height[right])*(right-left);res=Math.max(res,area);if(height[right]>=height[left]){left++;}else{right--;}}return res;}
}

15. 三数之和

https://leetcode.cn/problems/3sum/

class Solution {public List<List<Integer>> threeSum(int[] nums) {int len=nums.length;Arrays.sort(nums);//先排序List<List<Integer>> res=new ArrayList<>();for(int i=0;i<len;i++){if(nums[i]>0){return res;}if(i>0 && nums[i]==nums[i-1]){continue;//去重}int left=i+1;int right=len-1;while(right>left){int sum=nums[i]+nums[left]+nums[right];if(sum<0){left++;}else if(sum>0){right--;}else{res.add(Arrays.asList(nums[i],nums[left],nums[right]));left++;right--;while(right>left && nums[left]==nums[left-1]){left++;}while(right>left && nums[right]==nums[right+1]){right--;}}}}return res;}
}

42. 接雨水*(暴力，双指针优化暴力，单调栈)

https://leetcode.cn/problems/trapping-rain-water/

class Solution {public int trap(int[] height) {//3.单调栈int len=height.length;if (len <= 2) return 0;int res=0;Stack<Integer> s=new Stack<>();s.push(0);for(int i=1;i<len;i++){int top=s.peek();if(height[top]>height[i]){s.push(i);}else if(height[top]>height[i]){s.pop();s.push(i);}else{int curHigh=height[i];//目前节点的高度while(!s.isEmpty() && (curHigh>height[s.peek()])){int mid=s.pop();if(!s.isEmpty()){int left=s.peek();int h=Math.min(height[left],curHigh)-height[mid];int w=i-left-1;int area=h*w;if(area>0){res+=area;}}}s.push(i);}}return res;//2.双指针/*为了得到两边的最高高度，使用了双指针来遍历，每到一个柱子都向两边遍历一遍，这其实是有重复计算的。我们把每一个位置的左边最高高度记录在一个数组上（maxLeft[]）右边最高高度记录在一个数组上（maxRight[]），这样就避免了重复计算。*/int len=height.length;if (len <= 2) return 0;int res=0;int maxLeft[]=new int[len];int maxRight[]=new int[len];// 记录每个柱子左边柱子最大高度maxLeft[0]=height[0];for(int i=1;i<len;i++){maxLeft[i]=Math.max(maxLeft[i-1],height[i]);}// 记录每个柱子右边柱子最大高度maxRight[len-1]=height[len-1];for(int i=len-2;i>=0;i--){maxRight[i]=Math.max(maxRight[i+1],height[i]);}for(int i=0;i<len;i++){if(i==0 || i==len-1){continue;}int h=Math.min(maxLeft[i],maxRight[i])-height[i];res+=h;}return res;//1.暴力解法//每一列雨水的高度，取决于该列左侧最高的柱子和右侧最高的柱子中最矮的//那个柱子的高度减去当前列的高度h=Math.min(maxLeft,maxRight)-height[i]int len=height.length;int res=0;for(int i=0;i<len;i++){if(i==0 || i==len-1){continue;}int maxLeft=0;int maxRight=0;for(int j=i;j<len;j++){maxRight=Math.max(maxRight,height[j]);}for(int j=i;j>=0;j--){maxLeft=Math.max(maxLeft,height[j]);}int h=Math.min(maxLeft,maxRight)-height[i];res+=h;}return res;}
}

27. 移除元素

https://leetcode.cn/problems/remove-element/description/

class Solution {public int removeElement(int[] nums, int val) {//1.快慢指针int slow=0;for(int fast=0;fast<nums.length;fast++){if(nums[fast]!=val){nums[slow++]=nums[fast];}}return slow;//2.相向双指针int leftIndex = 0;int rightIndex = nums.length - 1;while (leftIndex <= rightIndex) {// 找左边等于val的元素while (leftIndex <= rightIndex && nums[leftIndex] != val){++leftIndex;}// 找右边不等于val的元素while (leftIndex <= rightIndex && nums[rightIndex] == val) {-- rightIndex;}// 将右边不等于val的元素覆盖左边等于val的元素if (leftIndex < rightIndex) {nums[leftIndex++] = nums[rightIndex--];}}return leftIndex;   // leftIndex一定指向了最终数组末尾的下一个元素}
}

344. 反转字符串

https://leetcode.cn/problems/reverse-string/description/

class Solution {public void reverseString(char[] s) {int len=s.length;int left=0;int rigtht=len-1;while(left<rigtht){char t=s[left];s[left++]=s[rigtht];s[rigtht--]=t;}}
}

54. 替换数字（第八期模拟笔试）

https://kamacoder.com/problempage.php?pid=1064

import java.util.Scanner;class Main {public static void main(String[] args) {Scanner in = new Scanner(System.in);String s = in.nextLine();StringBuilder sb = new StringBuilder();for (int i = 0; i < s.length(); i++) {if (Character.isDigit(s.charAt(i))) {sb.append("number");}else sb.append(s.charAt(i));}System.out.println(sb);}
}

151. 反转字符串中的单词

https://leetcode.cn/problems/reverse-words-in-a-string/description/

class Solution {public String reverseWords(String s) {//去掉字符串中的多余空格StringBuilder sb=clearSpace(s);//对整个字符串进行翻转reverseAll(sb,0,sb.length()-1);//对每个单词进行翻转reverseStr(sb);return sb.toString();}//去掉字符串中的多余空格public StringBuilder clearSpace(String s){int len=s.length();int start=0;int end=len-1;while(s.charAt(start)==' ')start++;while(s.charAt(end)==' ')end--;StringBuilder sb=new StringBuilder();while(start<=end){char c=s.charAt(start);if(c!=' ' || sb.charAt(sb.length()-1)!=' '){sb.append(c);}start++;}return sb;}//对整个字符串进行翻转public void reverseAll(StringBuilder sb,int start,int end){while(start<end){char temp=sb.charAt(start);sb.setCharAt(start++,sb.charAt(end));sb.setCharAt(end--,temp);}}//对每个单词进行翻转public void reverseStr(StringBuilder sb){int start = 0;int end = 1;int n = sb.length();while (start < n) {while (end < n && sb.charAt(end) != ' ') {end++;}reverseAll(sb, start, end - 1);start = end + 1;end = start + 1;}}
}

206. 反转链表

https://leetcode.cn/problems/reverse-linked-list/description/

/*** 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 reverseList(ListNode head) {ListNode cur=head;ListNode pre=null;ListNode temp=null;while(cur!=null){temp=cur.next;cur.next=pre;pre=cur;cur=temp;}return pre; }
}

19. 删除链表的倒数第 N 个结点.

https://leetcode.cn/problems/remove-nth-node-from-end-of-list/description/

/*** 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 removeNthFromEnd(ListNode head, int n) {//2.双指针，快慢指针//如果要删除倒数第n个节点，让fast移动n步，然后让fast和slow同时移动，//fast和slow中间永远差n个节点，直到fast指向链表末尾。//删掉slow所指向的节点就可以了。ListNode preHead=new ListNode(0);preHead.next=head;ListNode slow=preHead;ListNode fast=preHead;while(n>0){fast=fast.next;n--;}fast=fast.next;while(fast!=null){fast=fast.next;slow=slow.next;}slow.next=slow.next.next;return preHead.next;//1.普通方法ListNode cur=head;int size=0;while(cur!=null){cur=cur.next;size++;}if(n==size){head=head.next;return head;}cur=head;for(int i=0;i<size-n-1;i++){cur=cur.next;}if(cur.next!=null)cur.next=cur.next.next;return head;}
}

面试题 02.07. 链表相交

https://leetcode.cn/problems/intersection-of-two-linked-lists-lcci/description/

交点不是数值相等，而是指针相等

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/
public class Solution {public ListNode getIntersectionNode(ListNode headA, ListNode headB) {ListNode pA=headA;ListNode pB=headB;int lenA=0;int lenB=0;while(pA!=null){pA=pA.next;lenA++;}while(pB!=null){pB=pB.next;lenB++;}pA=headA;pB=headB;if(lenA>lenB){int gap=lenA-lenB;while(gap>0){pA=pA.next;gap--;}while(pA!=null){if(pA==pB){return pA;}pA=pA.next;pB=pB.next;}}else{int gap=lenB-lenA;while(gap>0){pB=pB.next;gap--;}while(pA!=null){if(pA==pB){return pA;}pA=pA.next;pB=pB.next;}}return null;}
}

142. 环形链表 II

https://leetcode.cn/problems/linked-list-cycle-ii/

/*** Definition for singly-linked list.* class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/
public class Solution {public ListNode detectCycle(ListNode head) {ListNode slow=head;ListNode fast=head;while(fast!=null && fast.next!=null){slow=slow.next;fast=fast.next.next;if(slow==fast){ListNode index01=head;ListNode index02=fast;while(index01!=index02){index01=index01.next;index02=index02.next;}return index01;}}return null;}
}