Day 27 | 39. Combination Sum | 40. Combination Sum II | 131. Palindrome Partitioning

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

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LeetCode 39. Combination Sum

Question Link

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    List<Integer> path = new ArrayList<>();
    int sum;

    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        Arrays.sort(candidates);
        backTracking(0, candidates, target);
        return result;
    }

    void backTracking(int index, int[] candidates, int target){
        if(sum == target){
            result.add(new ArrayList<>(path));
            return;
        }
        if(sum > target)
            return;
        for(int i = index; i < candidates.length && sum + candidates[i] <= target; i++){
            sum += candidates[i];
            path.add(candidates[i]);
            backTracking(i, candidates, target);
            sum -= candidates[i];
            path.remove(path.size() - 1);  
        }
    }
}

• Sort the array first.
• If the sum of the next loop(sum+candidates[i]) is larger than the target, stop the looping.

LeetCode 40. Combination Sum II

Question Link

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    List<Integer> path = new ArrayList<>();
    int sum;
    boolean[] used;

    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        used = new boolean[candidates.length];
        Arrays.sort(candidates);
        backTracing(0, candidates, target);
        return result;
    }

    void backTracing(int index, int[] candidates, int target){
        if(sum == target){
            result.add(new ArrayList<>(path));
            return;
        }
        if(sum > target)
            return;

        for(int i = index; i < candidates.length; i++){
            // deduplication
            if(i > 0 && candidates[i] == candidates[i-1] && used[i-1]==false) 
                continue;
            sum += candidates[i];
            path.add(candidates[i]);
            used[i] = true;
            backTracing(i+1, candidates, target);
            sum -= candidates[i];
            path.remove(path.size()-1);
            used[i] = false;
        }
    }
}

• We need to deduplicate the candidates.
• When candidates[i] == candidates[i - 1]：
  • If used[i - 1] == true, it demonstrates that candidates[i - 1] was used in the same leaf.
  • If used[i - 1] == false, it demonstrates that candidates[i - 1] was used in the same layer.

LeetCode 131. Palindrome Partitioning

Question Link

class Solution {
    List<List<String>> result = new ArrayList<>();
    List<String> path = new ArrayList<>();

    public List<List<String>> partition(String s) {
        backTracking(0, s);
        return result;
    }

    void backTracking(int index, String s){
        if(index >= s.length()){
            result.add(new ArrayList<>(path));
            return;
        }
        for(int i = index; i < s.length(); i++){
            if(isPalindrome(s, index, i)){
                // substring [startIndex, endIndex)
                String str = s.substring(index, i+1);
                path.add(str);
            } else
                continue;
            backTracking(i+1, s);
            path.remove(path.size() - 1);
        }
    }

    boolean isPalindrome(String s, int start, int end){
        while(start < end){
            if(s.charAt(start) != s.charAt(end))
                return false;
            start++;
            end--;
        }
        return true;
    }
}

• substring(startIndex, endIndex) includes the element in startIndex but doesn’t includes the element in the endIndex, which is [startIndex, endIndex)
• Use the Double Pointer Method to determine the palindrome