力扣labuladong一刷day2共8题 | 26. 删除有序数组中的重复项 83. 删除排序链表中的重复元素

一、26. 删除有序数组中的重复项

题目链接：https://leetcode.cn/problems/remove-duplicates-from-sorted-array/?utm_source=LCUS&utm_medium=ip_redirect&utm_campaign=transfer2china
思路：快慢指针，因为要删除重复元素，慢指针只有在快指针与前一个元素不等时才往前走一步，收集一个唯一元素。

class Solution {public int removeDuplicates(int[] nums) {if (nums.length == 1) return 1;int slow = 0, k = 1;for (int i = 1; i < nums.length; i++) {if (nums[i] != nums[i-1]) {k++;nums[++slow] = nums[i];}}return k;}
}

二、83. 删除排序链表中的重复元素

题目链接：https://leetcode.cn/problems/remove-duplicates-from-sorted-list/
思路：和上一题类似，也是快慢指针，只不过是在链表里操作。

class Solution {public ListNode deleteDuplicates(ListNode head) {if (head == null) return null;ListNode root = new ListNode(-1, head);ListNode slow = head, fast = head;while (fast != null) {if (slow.val != fast.val) {slow.next = fast;slow = slow.next;}fast = fast.next;}slow.next = null;return root.next;}
}

三、27. 移除元素

题目链接：https://leetcode.com/problems/remove-element/
思路：快慢指针，快指针元素与要删除的元素不相等就用快指针覆盖慢指针，如果相等就只需要快指针前进。

class Solution {public int removeElement(int[] nums, int val) {if (nums.length == 0) return 0;int slow = 0;for (int i = 0; i < nums.length; i++) {if (nums[i] != val) {nums[slow++] = nums[i];}}return slow;}
}

四、283. 移动零

题目链接：https://leetcode.cn/problems/move-zeroes/
思路：和前面的删除元素是一样的，只不过把0都删除以后再把数组后面的元素都改成0.

class Solution {public void moveZeroes(int[] nums) {int slow = 0;for (int i = 0; i < nums.length; i++) {if (nums[i] != 0) {nums[slow++] = nums[i];}}for (int i = slow; i < nums.length; i++) {nums[i] = 0;}}
}

五、167. 两数之和 II - 输入有序数组

题目链接：https://leetcode.cn/problems/two-sum-ii-input-array-is-sorted/
思路：要求两数之和等于target，可以采用类似于二分查找的方法，left=0，right=nums.length-1.sum>tagert right–，sum<target left++；

class Solution {public int[] twoSum(int[] numbers, int target) {int left = 0, right = numbers.length-1;while (left < right) {int sum = numbers[left] + numbers[right];if (sum == target) {return new int[]{left+1, right+1};}else if (sum < target) {left++;}else {right--;}}return new int[]{-1, -1};}
}

六、344. 反转字符串

题目链接：https://leetcode.cn/problems/reverse-string/
思路：左右指针。

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

七、5. 最长回文子串

题目链接：https://leetcode.cn/problems/longest-palindromic-substring/
思路：寻找最长的回文子串，利用回文子串的特性，应该基于任意一个点从中间向两端扩散寻找，扩散时分为奇数和偶数进行扩散。

class Solution {public String longestPalindrome(String s) {String res = "";for (int i = 0; i < s.length(); i++) {String s1 = f1(s, i, i);String s2 = f1(s, i, i+1);res = res.length() > s1.length() ? res : s1;res = res.length() > s2.length() ? res : s2;}return res;}String f1(String s, int l, int r) {while (l >= 0 && r <= s.length()-1 && s.charAt(l) == s.charAt(r)) {l--;r++;}return s.substring(l+1, r);}
}