first

Author: joker53-1

src/problem/mod.rs

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+mod p0001_two_sum;
+mod p0002_add_two_numbers;
+mod p0003_longest_substring_without_repeating_characters;
+mod p0004_median_of_two_sorted_arrays;
+mod p0005_longest_palindromic_substring;
+mod p0006_zigzag_conversion;
+mod p0007_reverse_integer;
+mod p0008_string_to_integer_atoi;
+mod p0009_palindrome_number;
+mod p0094_binary_tree_inorder_traversal;
+mod p0095_unique_binary_search_trees_ii;
+mod p0096_unique_binary_search_trees;
+mod p0098_validate_binary_search_tree;

src/problem/p0001_two_sum.rs

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+use std::collections::HashMap;
+
+/**
+ * [1] Two Sum
+ *
+ * Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
+ * You may assume that each input would have exactly one solution, and you may not use the same element twice.
+ * You can return the answer in any order.
+ *  
+ * <strong class="example">Example 1:
+ * 
+ * Input: nums = [2,7,11,15], target = 9
+ * Output: [0,1]
+ * Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
+ * 
+ * <strong class="example">Example 2:
+ * 
+ * Input: nums = [3,2,4], target = 6
+ * Output: [1,2]
+ * 
+ * <strong class="example">Example 3:
+ * 
+ * Input: nums = [3,3], target = 6
+ * Output: [0,1]
+ * 
+ *  
+ * Constraints:
+ * 
+ * 	2 <= nums.length <= 10^4
+ * 	-10^9 <= nums[i] <= 10^9
+ * 	-10^9 <= target <= 10^9
+ * 	Only one valid answer exists.
+ * 
+ *  
+ * Follow-up: Can you come up with an algorithm that is less than O(n^2)<font face="monospace"> </font>time complexity?
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.com/problems/two-sum/
+// discuss: https://leetcode.com/problems/two-sum/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> {
+        let mut map = HashMap::new();
+        nums.iter().enumerate().for_each(|(i, &v)| {
+            map.insert(target-v, i);
+        });
+        for (idx, x) in nums.iter().enumerate() {
+            if let Some(&i) = map.get(&(x)) {
+                if i != idx {
+                    return vec![idx as i32, i as i32];
+                }
+            }
+        }
+        vec![]
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_1() {
+    }
+}

src/problem/p0002_add_two_numbers.rs

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+/**
+ * [2] Add Two Numbers
+ *
+ * You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.
+ * You may assume the two numbers do not contain any leading zero, except the number 0 itself.
+ *  
+ * <strong class="example">Example 1:
+ * <img alt="" src="https://assets.leetcode.com/uploads/2020/10/02/addtwonumber1.jpg" style="width: 483px; height: 342px;" />
+ * Input: l1 = [2,4,3], l2 = [5,6,4]
+ * Output: [7,0,8]
+ * Explanation: 342 + 465 = 807.
+ * 
+ * <strong class="example">Example 2:
+ * 
+ * Input: l1 = [0], l2 = [0]
+ * Output: [0]
+ * 
+ * <strong class="example">Example 3:
+ * 
+ * Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
+ * Output: [8,9,9,9,0,0,0,1]
+ * 
+ *  
+ * Constraints:
+ * 
+ * 	The number of nodes in each linked list is in the range [1, 100].
+ * 	0 <= Node.val <= 9
+ * 	It is guaranteed that the list represents a number that does not have leading zeros.
+ * 
+ */
+pub struct Solution {}
+use crate::util::linked_list::{ListNode, to_list};
+
+// problem: https://leetcode.com/problems/add-two-numbers/
+// discuss: https://leetcode.com/problems/add-two-numbers/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+// Definition for singly-linked list.
+// #[derive(PartialEq, Eq, Clone, Debug)]
+// pub struct ListNode {
+//   pub val: i32,
+//   pub next: Option<Box<ListNode>>
+// }
+// 
+// impl ListNode {
+//   #[inline]
+//   fn new(val: i32) -> Self {
+//     ListNode {
+//       next: None,
+//       val
+//     }
+//   }
+// }
+impl Solution {
+    pub fn add_two_numbers(l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
+        let (mut l1, mut l2) = (l1, l2);
+        let mut head = ListNode::new(0);
+        let mut tail = &mut head;
+        let mut carry = 0;
+        while l1.is_some() || l2.is_some() {
+            if let Some(node) = l1{
+                carry += node.val;
+                l1 = node.next;
+            }
+            if let Some(node) = l2{
+                carry += node.val;
+                l2 = node.next;
+            }
+            tail.next = Some(Box::new(ListNode::new(carry % 10)));
+
+            carry = carry / 10;
+            tail = tail.next.as_mut()?;
+        }
+        if carry > 0 {
+            tail.next = Some(Box::new(ListNode::new(carry)));
+        }
+        head.next
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_2() {
+    }
+}

src/problem/p0003_longest_substring_without_repeating_characters.rs

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+use std::cmp::max;
+use std::collections::{HashMap, HashSet};
+
+/**
+ * [3] Longest Substring Without Repeating Characters
+ *
+ * Given a string s, find the length of the longest <span data-keyword="substring-nonempty">substring</span> without duplicate characters.
+ *  
+ * <strong class="example">Example 1:
+ * 
+ * Input: s = "abcabcbb"
+ * Output: 3
+ * Explanation: The answer is "abc", with the length of 3.
+ * 
+ * <strong class="example">Example 2:
+ * 
+ * Input: s = "bbbbb"
+ * Output: 1
+ * Explanation: The answer is "b", with the length of 1.
+ * 
+ * <strong class="example">Example 3:
+ * 
+ * Input: s = "pwwkew"
+ * Output: 3
+ * Explanation: The answer is "wke", with the length of 3.
+ * Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.
+ * 
+ *  
+ * Constraints:
+ * 
+ * 	0 <= s.length <= 5 * 10^4
+ * 	s consists of English letters, digits, symbols and spaces.
+ * 
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.com/problems/longest-substring-without-repeating-characters/
+// discuss: https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn length_of_longest_substring(s: String) -> i32 {
+        let chars = s.chars().collect::<Vec<char>>();
+        let mut map = HashMap::new();
+        let mut ans = 0;
+        let mut left = 0;
+        for i in 0..chars.len() {
+            if map.contains_key(&chars[i]) {
+                left = max(map[&chars[i]] + 1, left);
+            }
+            map.insert(chars[i], i);
+            ans = max(ans, i - left + 1);
+        }
+        ans as i32
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_3() {
+        println!("{}", Solution::length_of_longest_substring("abba".to_string()));
+    }
+}

src/problem/p0004_median_of_two_sorted_arrays.rs

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+/**
+ * [4] Median of Two Sorted Arrays
+ *
+ * Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.
+ * The overall run time complexity should be O(log (m+n)).
+ *  
+ * <strong class="example">Example 1:
+ *
+ * Input: nums1 = [1,3], nums2 = [2]
+ * Output: 2.00000
+ * Explanation: merged array = [1,2,3] and median is 2.
+ *
+ * <strong class="example">Example 2:
+ *
+ * Input: nums1 = [1,2], nums2 = [3,4]
+ * Output: 2.50000
+ * Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
+ *
+ *  
+ * Constraints:
+ *
+ * 	nums1.length == m
+ * 	nums2.length == n
+ * 	0 <= m <= 1000
+ * 	0 <= n <= 1000
+ * 	1 <= m + n <= 2000
+ * 	-10^6 <= nums1[i], nums2[i] <= 10^6
+ *
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.com/problems/median-of-two-sorted-arrays/
+// discuss: https://leetcode.com/problems/median-of-two-sorted-arrays/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn find_median_sorted_arrays(nums1: Vec<i32>, nums2: Vec<i32>) -> f64 {
+        let size = nums1.len() + nums2.len();
+        if size == 1 { return if nums1.len() == 1 { nums1[0] as f64 } else { nums2[0] as f64 }  }
+        let mid = size / 2;
+        let (mut n1, mut n2, mut i) = (0, 0, 0);
+        let mut pre = -1;
+        let mut cur = -1;
+        while i <= mid && n1 < nums1.len() && n2 < nums2.len() {
+            pre = cur;
+            if nums1[n1] < nums2[n2] {
+                cur = nums1[n1];
+                n1 += 1
+            } else {
+                cur = nums2[n2];
+                n2 += 1
+            }
+            i += 1;
+        }
+        while i <= mid && n1 < nums1.len() {
+            pre = cur;
+            cur = nums1[n1];
+            n1 += 1;
+            i += 1;
+        }
+        while i <= mid && n2 < nums2.len() {
+            pre = cur;
+            cur = nums2[n2];
+            n2 += 1;
+            i += 1;
+        }
+        if size % 2 == 0 {  (pre + cur) as f64/2.0  } else { cur as f64 }
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_4() {}
+}