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
Day 36 | 435. Non-overlapping Intervals | 763. Partition Labels | 56. Merge Intervals
Day 37 | 738. Monotone Increasing Digits | 714. Best Time to Buy and Sell Stock | 968. BT Camera
Day 38 | 509. Fibonacci Number | 70. Climbing Stairs | 746. Min Cost Climbing Stairs
Day 39 | 62. Unique Paths | 63. Unique Paths II
Day 41 | 343. Integer Break | 96. Unique Binary Search Trees
Day 42 | 0-1 Backpack Basic Theory（一）| 0-1 Backpack Basic Theory（二）| 416. Partition Equal Subset Sum
Day 43 | 1049. Last Stone Weight II | 494. Target Sum | 474. Ones and Zeroes
Day 44 | Full Backpack Basic Theory | 518. Coin Change II | 377. Combination Sum IV
Day 45 | 70. Climbing Stairs | 322. Coin Change | 279. Perfect Squares

Directory

• 139. Word Break
• Backpack Question Summary
• • • Recursion Formula
    • Traversal Order

139. Word Break

Question Link

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;

        for(int i = 1; i <= s.length(); i++){
            for(String word : wordDict){
                int len = word.length();
                // substring(beginIndex, endIndex) return the content from beginIndex to endIndex-1
                if(i >= len && dp[i-len] && word.equals(s.substring(i-len, i)))
                    dp[i] = true;
            }
        }

        return dp[s.length()];
    }
}

• dp[i]: demonstrate whether a string of length i could be segmented into one or more dictionary words.

• dp[0] = true, cause dp[0] is the root of the recursion. Otherwise, all the following recursion will be false.

• s.substring(beginIndex, endIndex) return the content from beginIndex to endIndex-1

Backpack Question Summary

Recursion Formula

• When asking whether the backpack can be filled(or how much it can hold at most)
  • dp[j] = max(dp[j], dp[j - nums[i]] + nums[i])
• When asking the number of method to fill the backpack
  • dp[j] += dp[j - nums[i]]
• When asking the maximum value of the backpack
  • dp[j] = max(dp[j], dp[j - weight[i]] + value[i])
• When asking the minimum number of items to fill the backpack
  • dp[j] = min(dp[j - coins[i]] + 1, dp[j])

Traversal Order

• 01 backpack
  • If we use one dimension array, we must traverse items first, and the inner loop must traverses from large to small.
• Full backpack
  • traverse items first and traverse capacity first are both fine.
  • The inner loop must traverses from small to large.
  • If we solve for the number of combinations, the outer loop traverses items, the inner loop traverses capacity.
  • If we solve for the number of permutations, the outer loop traverses capacity, the inner loop traverses items.