力扣26 快慢指针 去掉重复元素 数组

 public int removeDuplicates(int[] nums) {int slow = 0;int fast = 1;int count = 0;if (nums.length == 1) {return 1;} else {while (fast < nums.length) {if (nums[slow] == nums[fast]) {// 若相等 则 fast向前走 slow不动fast++;}else{//  若不相等 则 slow后一位设置为fast位置的元素 fast向后走slow++;nums[slow] = nums[fast];fast++;}}}return slow+1;}

力扣83 快慢指针向后遍历 去掉重复元素（链表）

 public ListNode deleteDuplicates(ListNode head) {if(head==null || head.next==null){return head;}else{// head 不为空 同时 head.next 不为空ListNode slow = head;ListNode fast = head.next;while(fast != null){if(slow.val == fast.val){slow.next = fast.next;fast = fast.next;}else{slow = fast;fast = fast.next;}}return head;}}

力扣 27 快慢指针向后遍历 去掉val元素

public int removeElement(int[] nums, int val) {int fast = 0, slow = 0;//如果 fast 遇到值为 val 的元素，则直接跳过，//否则就赋值给 slow 指针，并让 slow 前进一步。while (fast < nums.length) {if (nums[fast] != val) {nums[slow] = nums[fast];slow++;}fast++;}// 前面slow++ 所以返回slowreturn slow;}

力扣283 快慢指针向后遍历 除零

public void moveZeroes(int[] nums) {int length = nums.length;int slow = 0, fast = 0;while(fast < length){if(nums[fast] == 0){// fast == 0 则跳过 fast向前进fast++;}else{// fast不为0 则 将fast当前的值送到左侧，slow和fast都向前走一步nums[slow] = nums[fast];slow++;fast++;}}for(int i=slow;i<length;i++){// 剩余部分赋零nums[i] = 0;}}

力扣 344 左右向中间简单交换

 public void reverseString(char[] s) {for (int i = 0; i < s.length / 2; i++) {char temp = s[i];s[i] = s[s.length - 1 - i];s[s.length - 1 - i] = temp;}}

力扣1 HashMap

public int[] twoSum(int[] nums, int target) {HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();for(int i=0;i<nums.length;i++){int need = target - nums[i];if(map.containsKey(need)){return new int[]{i,map.get(need)};}else{map.put(nums[i],i);}}return new int[]{};}