力扣爆刷第112天之CodeTop100五连刷46-50

一、148. 排序链表

题目链接：https://leetcode.cn/problems/sort-list/description/
思路：链表的归并排序，先递归分割链表，然后比较合并，注意分割操作，和边界条件。

class Solution {public ListNode sortList(ListNode head) {return cutList(head, null);}ListNode cutList(ListNode head, ListNode end) {if(head == null) return head;if(head.next == end) {head.next = null;return head;}ListNode slow = head, fast = head;while(fast != end) {slow = slow.next;fast = fast.next;if(fast != end) {fast = fast.next;}}ListNode left = cutList(head, slow);ListNode right = cutList(slow, end);return merge(left, right);}ListNode merge(ListNode node1, ListNode node2) {ListNode root = new ListNode(), temp = root;while(node1 != null && node2 != null) {if(node1.val <= node2.val) {temp.next = node1;node1 = node1.next;}else{temp.next = node2;node2 = node2.next;}temp = temp.next;}if(node1 != null) {temp.next = node1;}if(node2 != null) {temp.next = node2;}return root.next;}}

二、22. 括号生成

题目链接：https://leetcode.cn/problems/generate-parentheses/description/
思路：排列题，但是不需要考虑去重，只需要正常排列，然后收获结果的时候进行判断是否合法。

class Solution {char[] source  = {'(', ')'};List<String> list = new ArrayList<>();StringBuilder builder = new StringBuilder();public List<String> generateParenthesis(int n) {backTracking(n*2);return list;}void backTracking(int n) {if(builder.length() == n) {if(isTrue(builder)) {list.add(builder.toString());}return;}for(int i = 0; i < 2; i++) {builder.append(source[i]);backTracking(n);builder.deleteCharAt(builder.length()-1);}}boolean isTrue(StringBuilder builder) {int n = 0;for (int i = 0; i < builder.length(); i++) {if(builder.charAt(i) == '(') {n++;}else{n--;}if(n < 0) return false;}return n == 0;}
}

三、70. 爬楼梯

题目链接：https://leetcode.cn/problems/climbing-stairs/description/
思路：最简单的动态规划，每一个位置依赖于前两个位置，分别为跨一步和跨两步。

class Solution {public int climbStairs(int n) {if(n < 4) return n;int a = 1, b = 2, c = 0;for(int i = 3; i <= n; i++) {c = a + b;a = b;b = c;}return c;}
}

四、2. 两数相加

题目链接：https://leetcode.cn/problems/add-two-numbers/description/
思路：遍历相加，然后用一个变量记录进制值，下次计算时加进去。

class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {int res = 0;ListNode root = new ListNode(), p = root;while(l1 != null || l2 != null) {int a = l1 != null ? l1.val : 0;int b = l2 != null ? l2.val : 0;int sum = a+b+res;res = sum / 10;ListNode t = new ListNode(sum%10);p.next = t;p = t;l1 = l1 != null ? l1.next : null;l2 = l2 != null ? l2.next : null;}    if(res == 1) {p.next = new ListNode(1);}return root.next;}}

五、165. 比较版本号

题目链接：https://leetcode.cn/problems/compare-version-numbers/description/
思路：利用乘10循环计数的方法，前导0乘10后还是0，剩下的就简单轻松了，不等就判断返回，等就进行下一轮计算。

class Solution {public int compareVersion(String version1, String version2) {int i = 0, j = 0, m = version1.length(), n = version2.length();while(i < m || j < n) {int x = 0;for(; i < m && version1.charAt(i) != '.'; i++) {x = x * 10 + version1.charAt(i) - '0';}i++;int y = 0;for(; j < n && version2.charAt(j) != '.'; j++) {y = y * 10 + version2.charAt(j) - '0';}j++;if(x != y) {return x > y ? 1 : -1;}}return 0;}
}