Day 36 | 435. Non-overlapping Intervals | 763. Partition Labels | 56. Merge Intervals

Day 1 | 704. Binary Search | 27. Remove Element | 35. Search Insert Position | 34. First and Last Position of Element in Sorted Array
Day 2 | 977. Squares of a Sorted Array | 209. Minimum Size Subarray Sum | 59. Spiral Matrix II
Day 3 | 203. Remove Linked List Elements | 707. Design Linked List | 206. Reverse Linked List
Day 4 | 24. Swap Nodes in Pairs| 19. Remove Nth Node From End of List| 160.Intersection of Two Lists
Day 6 | 242. Valid Anagram | 349. Intersection of Two Arrays | 202. Happy Numbe | 1. Two Sum
Day 7 | 454. 4Sum II | 383. Ransom Note | 15. 3Sum | 18. 4Sum
Day 8 | 344. Reverse String | 541. Reverse String II | 替换空格 | 151.Reverse Words in a String | 左旋转字符串
Day 9 | 28. Find the Index of the First Occurrence in a String | 459. Repeated Substring Pattern
Day 10 | 232. Implement Queue using Stacks | 225. Implement Stack using Queue
Day 11 | 20. Valid Parentheses | 1047. Remove All Adjacent Duplicates In String | 150. Evaluate RPN
Day 13 | 239. Sliding Window Maximum | 347. Top K Frequent Elements
Day 14 | 144.Binary Tree Preorder Traversal | 94.Binary Tree Inorder Traversal| 145.Binary Tree Postorder Traversal
Day 15 | 102. Binary Tree Level Order Traversal | 226. Invert Binary Tree | 101. Symmetric Tree
Day 16 | 104.MaximumDepth of BinaryTree| 111.MinimumDepth of BinaryTree| 222.CountComplete TreeNodes
Day 17 | 110. Balanced Binary Tree | 257. Binary Tree Paths | 404. Sum of Left Leaves
Day 18 | 513. Find Bottom Left Tree Value | 112. Path Sum | 105&106. Construct Binary Tree
Day 20 | 654. Maximum Binary Tree | 617. Merge Two Binary Trees | 700.Search in a Binary Search Tree
Day 21 | 530. Minimum Absolute Difference in BST | 501. Find Mode in Binary Search Tree | 236. Lowes
Day 22 | 235. Lowest Common Ancestor of a BST | 701. Insert into a BST | 450. Delete Node in a BST
Day 23 | 669. Trim a BST | 108. Convert Sorted Array to BST | 538. Convert BST to Greater Tree
Day 24 | 77. Combinations
Day 25 | 216. Combination Sum III | 17. Letter Combinations of a Phone Number
Day 27 | 39. Combination Sum | 40. Combination Sum II | 131. Palindrome Partitioning
Day 28 | 93. Restore IP Addresses | 78. Subsets | 90. Subsets II
Day 29 | 491. Non-decreasing Subsequences | 46. Permutations | 47. Permutations II
Day 30 | 332. Reconstruct Itinerary | 51. N-Queens | 37. Sudoku Solver
Day 31 | 455. Assign Cookies | 376. Wiggle Subsequence | 53. Maximum Subarray
Day 32 | 122. Best Time to Buy and Sell Stock II | 55. Jump Game | 45. Jump Game II
Day 34 | 1005. Maximize Sum Of Array After K Negations | 134. Gas Station | 135. Candy
Day 35 | 860. Lemonade Change | 406. Queue Reconstruction by Height | 452. Minimum Number of Arrows

Directory

  • LeetCode 435. Non-overlapping Intervals
  • LeetCode 763. Partition Labels
  • LeetCode 56. Merge Intervals


LeetCode 435. Non-overlapping Intervals

Question Link

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> {
            return Integer.compare(a[0], b[0]);
        });
        int count = 1;  // The number of non-overlapping intervals
        for(int i = 1; i < intervals.length; i++){
            if(intervals[i][0] < intervals[i-1][1])
                intervals[i][1] = Math.min(intervals[i-1][1], intervals[i][1]); // Update the right border of interval i
            else
                count++; 
        }
        return intervals.length - count;     // The number of intervals need to be removed
    }
}
  • This question is similar to 452. Minimum Number of Arrows to Burst Balloons.
  • The number of arrows equals the number of non-overlapping intervals.
  • Use the total number of intervals minus the number of arrows. We can get how many intervals we should remove.

LeetCode 763. Partition Labels

Question Link

class Solution {
    public List<Integer> partitionLabels(String s) {
        List<Integer> list = new LinkedList<>();
        int[] edge = new int[26];
        char[] chars = s.toCharArray();
        // Count the last present position of every character
        for(int i = 0; i < chars.length; i++)
            edge[chars[i] - 'a'] = i;
        
        int idx = 0;
        int last = -1;
        for(int i = 0; i < chars.length; i++){
            // Update the index of the furthest occurrence of the character
            idx = Math.max(idx, edge[chars[i] - 'a']);
            // If we find a character whose index is equal to the index of the furthest occurrence character, it is the split point.
            if(i == idx){
                list.add(i - last);
                last = i;
            }
        }
        return list;
    }
}
  • Count the last present position of every character
  • Traverse the characters, and update the index of the furthest occurrence of the character. If we find a character whose index is equal to the index of the furthest occurrence character, it is the split point.

LeetCode 56. Merge Intervals

Question Link

class Solution {
    public int[][] merge(int[][] intervals) {
        // Sort by the left border ASC
        Arrays.sort(intervals, (o1, o2) -> Integer.compare(o1[0], o2[0]));

        LinkedList<int[]> res = new LinkedList<>();
        res.add(intervals[0]);
        for (int i = 1; i < intervals.length; i++) {
            // Merge
            if (intervals[i][0] <= res.getLast()[1]) {
                int start = res.getLast()[0];
                int end = Math.max(intervals[i][1], res.getLast()[1]);
                res.removeLast();
                res.add(new int[]{start, end});
            }
            else {
                res.add(intervals[i]);
            }         
        }
        return res.toArray(new int[res.size()][]);
    }
}

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