public class Solution {
    //快慢指針法
    public boolean hasCycle(ListNode head) {
        ListNode fast = head;
        ListNode low = head;
        while(fast != null && fast.next != null){
            fast = fast.next.next;
            low = low.next;
            //此時(shí)相遇了
            if(fast == low){
                return true;
            }
        }
        return false;
    }
}

  //快慢指針法
    public ListNode middleNode(ListNode head) {
        ListNode low = head,fast = head;
        while(fast != null && fast.next != null){
            //慢指針走一步
            low = low.next;
            //快指針走兩步
            fast = fast.next.next;
        }
        //奇數(shù)，fast.next = null時(shí)，low即為所求
        //偶數(shù)，fsat == null時(shí)，low即為所求
        return low;
    }

 public ListNode oddEvenList(ListNode head) {
        if(head == null){
            return head;
        }
        ListNode fastHead = head.next;
        ListNode lowTail = head;//奇數(shù)鏈表
        ListNode fastTail = fastHead;//偶數(shù)鏈表
        while(fastTail != null && fastTail.next != null){
            //更新奇數(shù)節(jié)點(diǎn)時(shí)，奇數(shù)節(jié)點(diǎn)的后一個(gè)節(jié)點(diǎn)需要指向偶數(shù)節(jié)點(diǎn)的后一個(gè)節(jié)點(diǎn)
            lowTail.next = fastTail.next;
            lowTail = lowTail.next;
           // 更新偶數(shù)節(jié)點(diǎn)時(shí)，偶數(shù)節(jié)點(diǎn)的后一個(gè)節(jié)點(diǎn)需要指向奇數(shù)節(jié)點(diǎn)的后一個(gè)節(jié)點(diǎn)
            fastTail.next = lowTail.next;
            fastTail = fastTail.next;
        }
        //合并兩個(gè)鏈表
        lowTail.next = fastHead;
        return head;
    }

 public ListNode deleteDuplicates(ListNode head) {
        //虛擬頭節(jié)點(diǎn)法
        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = head;
        ListNode prev = dummyHead;
        ListNode cur = prev.next;
        while(cur != null){
            //引入next指針
            ListNode next = cur.next;
            if(next == null){
                //只有一個(gè)元素
                return dummyHead.next;
            }
            if(cur.val != next.val){
                //cur不是重復(fù)節(jié)點(diǎn)，三指針都移動(dòng)
                cur = cur.next;
                prev = prev.next;
            }else{
                //讓next指針一直向后移動(dòng)，直到與cur.val不相等的節(jié)點(diǎn)位置
                while(next != null && cur.val == next.val){
                    next = next.next;
                }
                // 此時(shí)next指向了第一個(gè)不重復(fù)的元素
                // 此時(shí)prev - next之間所有重復(fù)元素全部刪除
                prev.next = next;
                cur = next;
            }
        }
        return dummyHead.next;
    }

 public List<List<Integer>> threeSum(int[] nums) {
        int n = nums.length;
        Arrays.sort(nums);
        List<List<Integer>> ans = new ArrayList<List<Integer>>();
        // 枚舉 a
        for (int first = 0; first < n; ++first) {
            // 需要和上一次枚舉的數(shù)不相同
            if (first > 0 && nums[first] == nums[first - 1]) {
                continue;
            }
            // c 對(duì)應(yīng)的指針初始指向數(shù)組的最右端
            int third = n - 1;
            int target = -nums[first];
            // 枚舉 b
            for (int second = first + 1; second < n; ++second) {
                // 需要和上一次枚舉的數(shù)不相同
                if (second > first + 1 && nums[second] == nums[second - 1]) {
                    continue;
                }
                // 需要保證 b 的指針在 c 的指針的左側(cè)
                while (second < third && nums[second] + nums[third] > target) {
                    --third;
                }
                // 如果指針重合，隨著 b 后續(xù)的增加
                // 就不會(huì)有滿足 a+b+c=0 并且 b<c 的 c 了，可以退出循環(huán)
                if (second == third) {
                    break;
                }
                if (nums[second] + nums[third] == target) {
                    List<Integer> list = new ArrayList<Integer>();
                    list.add(nums[first]);
                    list.add(nums[second]);
                    list.add(nums[third]);
                    ans.add(list);
                }
            }
        }
        return ans;
    }

 public ListNode partition(ListNode head, int x) {
        // 創(chuàng)建small和big兩個(gè)小鏈表的頭節(jié)點(diǎn)
        ListNode smallHead = new ListNode(-1);
        ListNode bigHead = new ListNode(-1);
        // 按照升序插入，因此需要尾插
        // 分別指向兩個(gè)子鏈表的尾部
        ListNode smallTail = smallHead;
        ListNode bigTail = bigHead;
        //遍歷原鏈表，根據(jù)值放入small和big鏈表中
        while(head != null){
            if(head.val < x){
                smallTail.next = head;
                smallTail = head;
            }else{
                bigTail.next = head;
                bigTail = head;
            }
            head = head.next;
        }
        bigTail.next = null;
        //拼接小鏈表和大鏈表
        smallTail.next = bigHead.next;
        return smallHead.next;
    }

public void merge(int[] nums1, int m, int[] nums2, int n) {
        int p1 = 0, p2 = 0;
        int[] sorted = new int[m + n];
        int cur;
        while (p1 < m || p2 < n) {
            if (p1 == m) {
                cur = nums2[p2++];
            } else if (p2 == n) {
                cur = nums1[p1++];
            } else if (nums1[p1] < nums2[p2]) {
                cur = nums1[p1++];
            } else {
                cur = nums2[p2++];
            }
            sorted[p1 + p2 - 1] = cur;
        }
        for (int i = 0; i != m + n; ++i) {
            nums1[i] = sorted[i];
        }
    }

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

//滑動(dòng)窗口法
 public int minSubArrayLen(int s, int[] nums) {
        int n = nums.length;
        if (n == 0) {
            return 0;
        }
        int ans = Integer.MAX_VALUE;
        int start = 0, end = 0;
        int sum = 0;
        while (end < n) {
            sum += nums[end];
            while (sum >= s) {
                ans = Math.min(ans, end - start + 1);
                sum -= nums[start];
                start++;
            }
            end++;
        }
        return ans == Integer.MAX_VALUE ? 0 : ans;
    }

public ListNode sortList(ListNode head) {
        if (head == null || head.next == null)
            return head;
        ListNode fast = head.next, slow = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode tmp = slow.next;
        slow.next = null;
        ListNode left = sortList(head);
        ListNode right = sortList(tmp);
        ListNode h = new ListNode(0);
        ListNode res = h;
        while (left != null && right != null) {
            if (left.val < right.val) {
                h.next = left;
                left = left.next;
            } else {
                h.next = right;
                right = right.next;
            }
            h = h.next;
        }
        h.next = left != null ? left : right;
        return res.next;
    }