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
Determine the meaning of dp[i][j]
dp[i][j]
is the maximum value of items that are taken from subscript 0
to i
and put into the backpack with capacity j
Determine the recursive formula
dp[i][j]
is dp[i-1][j]
. The item i
can’t be put into the backpack when the weight of item i
is greater than the backpack j
weight. So the value of items in the backpack remains the same as before.dp[i - 1][j - weight[i]]
is the maximum value when the capacity of backpack is j - weight[i]
and doesn’t contain the item i
. value[i]
is the value of item i
. So dp[i-1][j-weight[i]] + value[i]
is the maximum value when containing the item i
.dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i])
;Initialize the dynamic programming array
0
, like dp[i][0]
. No matter which items are selected, the total value must be 0
.j < weight[0]
, dp[0][j]
should be 0
.j >= weight[0]
, dp[0][j]
should be value[0]
dp[i][j]
is derived from the upper left value. So the other subscripts can be initialized to whatever values, because they will be overwritten. It’s more convenient to initialize the dynamic programming array to 0
.Traversal order
traversing item first
and traversing backpack capacity first
all work. The traversing item first
is better understood.// The size of the weight array is the number of items
for(int i = 1; i < weight.size(); i++) { // Traverse items
for(int j = 0; j <= bagweight; j++) { // Traverse backpack capacity
if (j < weight[i]) dp[i][j] = dp[i - 1][j];
else dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
}
Determine the meaning of dp[j]
dp[j]
is the maximum value of items carried in the backpack with capacity j
Determine the recursive formula
dp[j]
is dp[i-1][j]
. The item i
can’t be put into the backpack when the weight of item i
is greater than the backpack j
weight. So the value of items in the backpack remains the same as before.dp[j - weight[i]]
denotes the maximum value of items carried by a backpack with capacity j - weight[i]
and doesn’t contain the item i
. value[i]
is the value of item i
. So dp[j-weight[i]] + value[i]
is the maximum value when containing the item i
.dp[j] = max(dp[j], dp[j - weight[i]] + value[i])
;Initialize the dynamic programming array
0
, like dp[0]
. No matter which items are selected, the total value must be 0
.0
. In this way, the dp array will get the maximum value when doing the recursion instead of being overwritten by initial value.Traversal order
2D
dp traversal, the backpack capacity is from small to large
, while for 1D
dp traversal, the backpack is from large to small
. The reverse order traversal is to ensure that item i
is only put in once, cause the positive order traversal will lead to item 0
be repeatedly added multiple times.dp[i][j]
are computed through the previous layerdp[i-1][j]
. So the dp[i][j]
of this layer won’t be overwritten.for
loops. We must adopt traversing item first
instead of traversing backpack capacity first
. Because if we adopt traversing backpack capacity first
, then only one item will be put in each dp[j]
(backpack).for(int i = 0; i < weight.size(); i++) { // Traversing items
for(int j = bagWeight; j >= weight[i]; j--) { // Traversing backpack capacity
dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
}
}
link
class Solution {
public boolean canPartition(int[] nums) {
if(nums==null || nums.length==0) return false;
int sum = 0;
for(int num : nums)
sum += num;
// if sum is odd number, two subsets can not be equal
if(sum%2 != 0) return false;
int target = sum / 2;
int[] dp = new int[target + 1];
for(int i = 0; i < nums.length; i++){ // Traversing items
for(int j = target; j >= nums[i]; j--) // Traversing capacity
dp[j] = Math.max(dp[j], dp[j-nums[i]] + nums[i]);
}
return dp[target] == target;
}
}
i
both are nums[i]