Data Structure and Algorithm

• 11. Container With Most Water

  Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of the line i is at (i, ai) and (i, 0). Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.

  Notice that you may not slant the container.

  Example 1:

  

  Input: height = [1,8,6,2,5,4,8,3,7]
  Output: 49
  Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
  

  Example 2:

  Input: height = [1,1]
  Output: 1
  

  Example 3:

  Input: height = [4,3,2,1,4]
  Output: 16
  

  Example 4:

  Input: height = [1,2,1]
  Output: 2
  

  Constraints:

  • n == height.length
  • 2 <= n <= 105
  • 0 <= height[i] <= 104

  class Solution {
      public int maxArea(int[] height) {
          int res = 0;
          for (int i = 0, j = height.length - 1; i < j; ) {
              int minHeight = height[i] < height[j] ? height[i++] : height[j--];
              res = Math.max(res, (j - i + 1) * minHeight); //plus 1
          }
          return res;
      }
  }
  

• 70. Climbing Stairs

  You are climbing a staircase. It takes n steps to reach the top.

  Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

  Example 1:

  Input: n = 2
  Output: 2
  Explanation: There are two ways to climb to the top.
  1. 1 step + 1 step
  2. 2 steps
  

  Example 2:

  Input: n = 3
  Output: 3
  Explanation: There are three ways to climb to the top.
  1. 1 step + 1 step + 1 step
  2. 1 step + 2 steps
  3. 2 steps + 1 step
  

  Constraints:

  • 1 <= n <= 45

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

• 15. 3Sum

  Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

  Notice that the solution set must not contain duplicate triplets.

  Example 1:

  Input: nums = [-1,0,1,2,-1,-4]
  Output: [[-1,-1,2],[-1,0,1]]
  

  Example 2:

  Input: nums = []
  Output: []
  

  Example 3:

  Input: nums = [0]
  Output: []
  

  Constraints:

  • 0 <= nums.length <= 3000
  • -105 <= nums[i] <= 105

  class Solution {
      public List<List<Integer>> threeSum(int[] nums) {
          List<List<Integer>> res = new ArrayList<>();
          Arrays.sort(nums);
          for (int i = 0; i < nums.length - 2 && nums[i] <= 0; i++) {
              if (i > 0 && nums[i] == nums[i - 1]) continue;
              for (int j = i + 1, k = nums.length - 1; j < k; ) {
                  int sum = nums[j] + nums[k];
                  if (j > i + 1 && nums[j] == nums[j - 1] || sum < -nums[i]) {
                      j++;
                  } else if (k < nums.length - 1 && nums[k] == nums[k + 1] || sum > -nums[i]) {
                      k--;
                  } else {
                      res.add(Arrays.asList(nums[i], nums[j], nums[k]));
                      j++;
                      k--;
                  }
              }
          }
          return res;
      }
  }
  

• 141. Linked List Cycle

  Given head, the head of a linked list, determine if the linked list has a cycle in it.

  There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to. Note that pos is not passed as a parameter.

  Return true if there is a cycle in the linked list. Otherwise, return false.

  Example 1:

  

  Input: head = [3,2,0,-4], pos = 1
  Output: true
  Explanation: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).
  

  Example 2:

  

  Input: head = [1,2], pos = 0
  Output: true
  Explanation: There is a cycle in the linked list, where the tail connects to the 0th node.
  

  Example 3:

  

  Input: head = [1], pos = -1
  Output: false
  Explanation: There is no cycle in the linked list.
  

  Constraints:

  • The number of the nodes in the list is in the range [0, 104].
  • -105 <= Node.val <= 105
  • pos is -1 or a valid index in the linked-list.

  public class Solution {
      public boolean hasCycle(ListNode head) {
          if (head == null || head.next == null) return false; 
          ListNode fast = head.next, slow = head;
          while (fast != null && fast.next != null) {
              if (fast == slow) return true;
              fast = fast.next.next;
              slow = slow.next;
          }
          return false;
      }
  }
  

• 142. Linked List Cycle II

  Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

  There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to. Note that pos is not passed as a parameter.

  Notice that you should not modify the linked list.

  Example 1:

  

  Input: head = [3,2,0,-4], pos = 1
  Output: tail connects to node index 1
  Explanation: There is a cycle in the linked list, where tail connects to the second node.
  

  Example 2:

  

  Input: head = [1,2], pos = 0
  Output: tail connects to node index 0
  Explanation: There is a cycle in the linked list, where tail connects to the first node.
  

  Example 3:

  

  Input: head = [1], pos = -1
  Output: no cycle
  Explanation: There is no cycle in the linked list.
  

  Constraints:

  • The number of the nodes in the list is in the range [0, 104].
  • -105 <= Node.val <= 105
  • pos is -1 or a valid index in the linked-list.

  Follow up: Can you solve it using O(1) (i.e. constant) memory?

  public class Solution {
      public ListNode detectCycle(ListNode head) {
          if (head == null || head.next == null) return null;
          ListNode fast = head, slow = head;
          while (fast != null && fast.next != null) {
              fast = fast.next.next;
              slow = slow.next;
              if (fast == slow) {
                  fast = head;
                  while (fast != slow) {
                      fast = fast.next;
                      slow = slow.next;
                  }
                  return fast;
              }
          }
          return null;
      }
  }
  

• 206. Reverse Linked List

  Given the head of a singly linked list, reverse the list, and return the reversed list.

  Example 1:

  

  Input: head = [1,2,3,4,5]
  Output: [5,4,3,2,1]
  

  Example 2:

  

  Input: head = [1,2]
  Output: [2,1]
  

  Example 3:

  Input: head = []
  Output: []
  

  Constraints:

  • The number of nodes in the list is the range [0, 5000].
  • -5000 <= Node.val <= 5000

  class Solution {
      public ListNode reverseList(ListNode head) {
          if (head == null || head.next == null) return head;
          ListNode res = new ListNode();
          while (head != null) {
              ListNode cur = head;
              head = head.next;
              cur.next = res.next;
              res.next = cur;
          }
          return res.next;
      }
  }
  

• 24. Swap Nodes in Pairs

  Given a linked list, swap every two adjacent nodes and return its head.

  Example 1:

  

  Input: head = [1,2,3,4]
  Output: [2,1,4,3]
  

  Example 2:

  Input: head = []
  Output: []
  

  Example 3:

  Input: head = [1]
  Output: [1]
  

  Constraints:

  • The number of nodes in the list is in the range [0, 100].
  • 0 <= Node.val <= 100

  Follow up: Can you solve the problem without modifying the values in the list's nodes? (i.e., Only nodes themselves may be changed.)

  class Solution {
      public ListNode swapPairs(ListNode head) {
          if (head == null || head.next == null) return head;
          ListNode res = new ListNode();
          ListNode cur = res;
          while (head != null && head.next != null) {
              ListNode a = head, b = head.next;
              a.next = b.next;
              b.next = a;
              cur.next = b;
              cur = a;
              head = cur.next;
          }
          return res.next;
      }
  }