diff --git a/lcci/16.26.Calculator/README.md b/lcci/16.26.Calculator/README.md index ee22dbedcdbdc..a80a2f4faa454 100644 --- a/lcci/16.26.Calculator/README.md +++ b/lcci/16.26.Calculator/README.md @@ -207,7 +207,7 @@ class Solution { var sign: Character = "+" var stk = [Int]() let sArray = Array(s) - + for i in 0.. [String] { var vis: [Int: Int] = [0: -1] var s = 0, mx = 0, k = 0 - + for i in 0.. [String] { var vis: [Int: Int] = [0: -1] var s = 0, mx = 0, k = 0 - + for i in 0.. @@ -59,20 +69,19 @@ class Solution: def spiralMatrix(self, m: int, n: int, head: Optional[ListNode]) -> List[List[int]]: ans = [[-1] * n for _ in range(m)] - i = j = p = 0 - dirs = [[0, 1], [1, 0], [0, -1], [-1, 0]] + i = j = k = 0 + dirs = (0, 1, 0, -1, 0) while 1: ans[i][j] = head.val head = head.next - if not head: + if head is None: break while 1: - x, y = i + dirs[p][0], j + dirs[p][1] - if x < 0 or y < 0 or x >= m or y >= n or ~ans[x][y]: - p = (p + 1) % 4 - else: + x, y = i + dirs[k], j + dirs[k + 1] + if 0 <= x < m and 0 <= y < n and ans[x][y] == -1: i, j = x, y break + k = (k + 1) % 4 return ans ``` @@ -90,11 +99,11 @@ class Solution: class Solution { public int[][] spiralMatrix(int m, int n, ListNode head) { int[][] ans = new int[m][n]; - for (int[] row : ans) { + for (var row : ans) { Arrays.fill(row, -1); } - int i = 0, j = 0, p = 0; - int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; + int i = 0, j = 0, k = 0; + final int[] dirs = {0, 1, 0, -1, 0}; while (true) { ans[i][j] = head.val; head = head.next; @@ -102,14 +111,13 @@ class Solution { break; } while (true) { - int x = i + dirs[p][0], y = j + dirs[p][1]; - if (x < 0 || y < 0 || x >= m || y >= n || ans[x][y] >= 0) { - p = (p + 1) % 4; - } else { + int x = i + dirs[k], y = j + dirs[k + 1]; + if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) { i = x; j = y; break; } + k = (k + 1) % 4; } } return ans; @@ -132,20 +140,22 @@ class Solution { public: vector> spiralMatrix(int m, int n, ListNode* head) { vector> ans(m, vector(n, -1)); - int i = 0, j = 0, p = 0; - vector> dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; + int i = 0, j = 0, k = 0; + const int dirs[5] = {0, 1, 0, -1, 0}; while (1) { ans[i][j] = head->val; head = head->next; - if (!head) break; + if (!head) { + break; + } while (1) { - int x = i + dirs[p][0], y = j + dirs[p][1]; - if (x < 0 || y < 0 || x >= m || y >= n || ans[x][y] >= 0) - p = (p + 1) % 4; - else { - i = x, j = y; + int x = i + dirs[k], y = j + dirs[k + 1]; + if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) { + i = x; + j = y; break; } + k = (k + 1) % 4; } } return ans; @@ -169,22 +179,20 @@ func spiralMatrix(m int, n int, head *ListNode) [][]int { ans[i][j] = -1 } } - i, j, p := 0, 0, 0 - dirs := [][]int{{0, 1}, {1, 0}, {0, -1}, {-1, 0}} + i, j, k := 0, 0, 0 + dirs := [5]int{0, 1, 0, -1, 0} for { ans[i][j] = head.Val - head = head.Next - if head == nil { + if head = head.Next; head == nil { break } for { - x, y := i+dirs[p][0], j+dirs[p][1] - if x < 0 || y < 0 || x >= m || y >= n || ans[x][y] >= 0 { - p = (p + 1) % 4 - } else { + x, y := i+dirs[k], j+dirs[k+1] + if x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1 { i, j = x, y break } + k = (k + 1) % 4 } } return ans @@ -205,26 +213,24 @@ func spiralMatrix(m int, n int, head *ListNode) [][]int { */ function spiralMatrix(m: number, n: number, head: ListNode | null): number[][] { - const dirs = [ - [0, 1], - [1, 0], - [0, -1], - [-1, 0], - ]; - let ans = Array.from({ length: m }, v => new Array(n).fill(-1)); - let i = 0, - j = 0, - k = 0; - while (head) { + const ans: number[][] = Array.from({ length: m }, () => Array(n).fill(-1)); + const dirs: number[] = [0, 1, 0, -1, 0]; + let [i, j, k] = [0, 0, 0]; + while (1) { ans[i][j] = head.val; head = head.next; - let x = i + dirs[k][0]; - let y = j + dirs[k][1]; - if (x < 0 || x > m - 1 || y < 0 || y > n - 1 || ans[x][y] != -1) { + if (!head) { + break; + } + while (1) { + const [x, y] = [i + dirs[k], j + dirs[k + 1]]; + if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] === -1) { + i = x; + j = y; + break; + } k = (k + 1) % 4; } - i = i + dirs[k][0]; - j = j + dirs[k][1]; } return ans; } diff --git a/solution/2300-2399/2326.Spiral Matrix IV/README_EN.md b/solution/2300-2399/2326.Spiral Matrix IV/README_EN.md index 3c45a58e5481b..ccae93c98083d 100644 --- a/solution/2300-2399/2326.Spiral Matrix IV/README_EN.md +++ b/solution/2300-2399/2326.Spiral Matrix IV/README_EN.md @@ -44,7 +44,17 @@ The last space in the matrix is set to -1. ## Solutions -### Solution 1 +### Solution 1: Simulation + +We define a two-dimensional array $\text{ans}$ to store the elements in the linked list, initially all filled with $-1$. We define three variables $i, j, k$, representing the current row, column, and direction respectively. We define an array $\text{dirs}$ to represent the offsets of the four directions. + +Then we start traversing the linked list. Each time we traverse a node, we fill the current node's value into $\text{ans}[i][j]$, then update the linked list pointer. If the linked list is empty, it means all elements have been filled and we exit the loop. + +Otherwise, we need to find the position of the next element. We can calculate the next position $(x, y)$ through the current position $(i, j)$ and the current direction $k$. If $(x, y)$ is within the range of the matrix, and $\text{ans}[x][y]$ is $-1$, it means $(x, y)$ has not been filled yet, so we take $(x, y)$ as the next position. Otherwise, we need to change the direction. + +After traversing the linked list, we get a spiral matrix and return it. + +The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$, where $m$ and $n$ represent the number of rows and columns of the matrix, respectively. @@ -57,20 +67,19 @@ The last space in the matrix is set to -1. class Solution: def spiralMatrix(self, m: int, n: int, head: Optional[ListNode]) -> List[List[int]]: ans = [[-1] * n for _ in range(m)] - i = j = p = 0 - dirs = [[0, 1], [1, 0], [0, -1], [-1, 0]] + i = j = k = 0 + dirs = (0, 1, 0, -1, 0) while 1: ans[i][j] = head.val head = head.next - if not head: + if head is None: break while 1: - x, y = i + dirs[p][0], j + dirs[p][1] - if x < 0 or y < 0 or x >= m or y >= n or ~ans[x][y]: - p = (p + 1) % 4 - else: + x, y = i + dirs[k], j + dirs[k + 1] + if 0 <= x < m and 0 <= y < n and ans[x][y] == -1: i, j = x, y break + k = (k + 1) % 4 return ans ``` @@ -88,11 +97,11 @@ class Solution: class Solution { public int[][] spiralMatrix(int m, int n, ListNode head) { int[][] ans = new int[m][n]; - for (int[] row : ans) { + for (var row : ans) { Arrays.fill(row, -1); } - int i = 0, j = 0, p = 0; - int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; + int i = 0, j = 0, k = 0; + final int[] dirs = {0, 1, 0, -1, 0}; while (true) { ans[i][j] = head.val; head = head.next; @@ -100,14 +109,13 @@ class Solution { break; } while (true) { - int x = i + dirs[p][0], y = j + dirs[p][1]; - if (x < 0 || y < 0 || x >= m || y >= n || ans[x][y] >= 0) { - p = (p + 1) % 4; - } else { + int x = i + dirs[k], y = j + dirs[k + 1]; + if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) { i = x; j = y; break; } + k = (k + 1) % 4; } } return ans; @@ -130,20 +138,22 @@ class Solution { public: vector> spiralMatrix(int m, int n, ListNode* head) { vector> ans(m, vector(n, -1)); - int i = 0, j = 0, p = 0; - vector> dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; + int i = 0, j = 0, k = 0; + const int dirs[5] = {0, 1, 0, -1, 0}; while (1) { ans[i][j] = head->val; head = head->next; - if (!head) break; + if (!head) { + break; + } while (1) { - int x = i + dirs[p][0], y = j + dirs[p][1]; - if (x < 0 || y < 0 || x >= m || y >= n || ans[x][y] >= 0) - p = (p + 1) % 4; - else { - i = x, j = y; + int x = i + dirs[k], y = j + dirs[k + 1]; + if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) { + i = x; + j = y; break; } + k = (k + 1) % 4; } } return ans; @@ -167,22 +177,20 @@ func spiralMatrix(m int, n int, head *ListNode) [][]int { ans[i][j] = -1 } } - i, j, p := 0, 0, 0 - dirs := [][]int{{0, 1}, {1, 0}, {0, -1}, {-1, 0}} + i, j, k := 0, 0, 0 + dirs := [5]int{0, 1, 0, -1, 0} for { ans[i][j] = head.Val - head = head.Next - if head == nil { + if head = head.Next; head == nil { break } for { - x, y := i+dirs[p][0], j+dirs[p][1] - if x < 0 || y < 0 || x >= m || y >= n || ans[x][y] >= 0 { - p = (p + 1) % 4 - } else { + x, y := i+dirs[k], j+dirs[k+1] + if x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1 { i, j = x, y break } + k = (k + 1) % 4 } } return ans @@ -203,26 +211,24 @@ func spiralMatrix(m int, n int, head *ListNode) [][]int { */ function spiralMatrix(m: number, n: number, head: ListNode | null): number[][] { - const dirs = [ - [0, 1], - [1, 0], - [0, -1], - [-1, 0], - ]; - let ans = Array.from({ length: m }, v => new Array(n).fill(-1)); - let i = 0, - j = 0, - k = 0; - while (head) { + const ans: number[][] = Array.from({ length: m }, () => Array(n).fill(-1)); + const dirs: number[] = [0, 1, 0, -1, 0]; + let [i, j, k] = [0, 0, 0]; + while (1) { ans[i][j] = head.val; head = head.next; - let x = i + dirs[k][0]; - let y = j + dirs[k][1]; - if (x < 0 || x > m - 1 || y < 0 || y > n - 1 || ans[x][y] != -1) { + if (!head) { + break; + } + while (1) { + const [x, y] = [i + dirs[k], j + dirs[k + 1]]; + if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] === -1) { + i = x; + j = y; + break; + } k = (k + 1) % 4; } - i = i + dirs[k][0]; - j = j + dirs[k][1]; } return ans; } diff --git a/solution/2300-2399/2326.Spiral Matrix IV/Solution.cpp b/solution/2300-2399/2326.Spiral Matrix IV/Solution.cpp index 1ae3cd31736c0..9de25fd390d18 100644 --- a/solution/2300-2399/2326.Spiral Matrix IV/Solution.cpp +++ b/solution/2300-2399/2326.Spiral Matrix IV/Solution.cpp @@ -12,20 +12,22 @@ class Solution { public: vector> spiralMatrix(int m, int n, ListNode* head) { vector> ans(m, vector(n, -1)); - int i = 0, j = 0, p = 0; - vector> dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; + int i = 0, j = 0, k = 0; + const int dirs[5] = {0, 1, 0, -1, 0}; while (1) { ans[i][j] = head->val; head = head->next; - if (!head) break; + if (!head) { + break; + } while (1) { - int x = i + dirs[p][0], y = j + dirs[p][1]; - if (x < 0 || y < 0 || x >= m || y >= n || ans[x][y] >= 0) - p = (p + 1) % 4; - else { - i = x, j = y; + int x = i + dirs[k], y = j + dirs[k + 1]; + if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) { + i = x; + j = y; break; } + k = (k + 1) % 4; } } return ans; diff --git a/solution/2300-2399/2326.Spiral Matrix IV/Solution.go b/solution/2300-2399/2326.Spiral Matrix IV/Solution.go index 325d971558b8f..f8a5901cfd99c 100644 --- a/solution/2300-2399/2326.Spiral Matrix IV/Solution.go +++ b/solution/2300-2399/2326.Spiral Matrix IV/Solution.go @@ -13,22 +13,20 @@ func spiralMatrix(m int, n int, head *ListNode) [][]int { ans[i][j] = -1 } } - i, j, p := 0, 0, 0 - dirs := [][]int{{0, 1}, {1, 0}, {0, -1}, {-1, 0}} + i, j, k := 0, 0, 0 + dirs := [5]int{0, 1, 0, -1, 0} for { ans[i][j] = head.Val - head = head.Next - if head == nil { + if head = head.Next; head == nil { break } for { - x, y := i+dirs[p][0], j+dirs[p][1] - if x < 0 || y < 0 || x >= m || y >= n || ans[x][y] >= 0 { - p = (p + 1) % 4 - } else { + x, y := i+dirs[k], j+dirs[k+1] + if x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1 { i, j = x, y break } + k = (k + 1) % 4 } } return ans diff --git a/solution/2300-2399/2326.Spiral Matrix IV/Solution.java b/solution/2300-2399/2326.Spiral Matrix IV/Solution.java index fff95f9618c10..10f3ea2f00704 100644 --- a/solution/2300-2399/2326.Spiral Matrix IV/Solution.java +++ b/solution/2300-2399/2326.Spiral Matrix IV/Solution.java @@ -11,11 +11,11 @@ class Solution { public int[][] spiralMatrix(int m, int n, ListNode head) { int[][] ans = new int[m][n]; - for (int[] row : ans) { + for (var row : ans) { Arrays.fill(row, -1); } - int i = 0, j = 0, p = 0; - int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; + int i = 0, j = 0, k = 0; + final int[] dirs = {0, 1, 0, -1, 0}; while (true) { ans[i][j] = head.val; head = head.next; @@ -23,14 +23,13 @@ public int[][] spiralMatrix(int m, int n, ListNode head) { break; } while (true) { - int x = i + dirs[p][0], y = j + dirs[p][1]; - if (x < 0 || y < 0 || x >= m || y >= n || ans[x][y] >= 0) { - p = (p + 1) % 4; - } else { + int x = i + dirs[k], y = j + dirs[k + 1]; + if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) { i = x; j = y; break; } + k = (k + 1) % 4; } } return ans; diff --git a/solution/2300-2399/2326.Spiral Matrix IV/Solution.py b/solution/2300-2399/2326.Spiral Matrix IV/Solution.py index 304faa37ce007..967535bc9a00a 100644 --- a/solution/2300-2399/2326.Spiral Matrix IV/Solution.py +++ b/solution/2300-2399/2326.Spiral Matrix IV/Solution.py @@ -6,18 +6,17 @@ class Solution: def spiralMatrix(self, m: int, n: int, head: Optional[ListNode]) -> List[List[int]]: ans = [[-1] * n for _ in range(m)] - i = j = p = 0 - dirs = [[0, 1], [1, 0], [0, -1], [-1, 0]] + i = j = k = 0 + dirs = (0, 1, 0, -1, 0) while 1: ans[i][j] = head.val head = head.next - if not head: + if head is None: break while 1: - x, y = i + dirs[p][0], j + dirs[p][1] - if x < 0 or y < 0 or x >= m or y >= n or ~ans[x][y]: - p = (p + 1) % 4 - else: + x, y = i + dirs[k], j + dirs[k + 1] + if 0 <= x < m and 0 <= y < n and ans[x][y] == -1: i, j = x, y break + k = (k + 1) % 4 return ans diff --git a/solution/2300-2399/2326.Spiral Matrix IV/Solution.ts b/solution/2300-2399/2326.Spiral Matrix IV/Solution.ts index 531a85dbee6f9..49ae43f342706 100644 --- a/solution/2300-2399/2326.Spiral Matrix IV/Solution.ts +++ b/solution/2300-2399/2326.Spiral Matrix IV/Solution.ts @@ -11,26 +11,24 @@ */ function spiralMatrix(m: number, n: number, head: ListNode | null): number[][] { - const dirs = [ - [0, 1], - [1, 0], - [0, -1], - [-1, 0], - ]; - let ans = Array.from({ length: m }, v => new Array(n).fill(-1)); - let i = 0, - j = 0, - k = 0; - while (head) { + const ans: number[][] = Array.from({ length: m }, () => Array(n).fill(-1)); + const dirs: number[] = [0, 1, 0, -1, 0]; + let [i, j, k] = [0, 0, 0]; + while (1) { ans[i][j] = head.val; head = head.next; - let x = i + dirs[k][0]; - let y = j + dirs[k][1]; - if (x < 0 || x > m - 1 || y < 0 || y > n - 1 || ans[x][y] != -1) { + if (!head) { + break; + } + while (1) { + const [x, y] = [i + dirs[k], j + dirs[k + 1]]; + if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] === -1) { + i = x; + j = y; + break; + } k = (k + 1) % 4; } - i = i + dirs[k][0]; - j = j + dirs[k][1]; } return ans; } diff --git a/solution/2300-2399/2391.Minimum Amount of Time to Collect Garbage/README.md b/solution/2300-2399/2391.Minimum Amount of Time to Collect Garbage/README.md index 081fd24930474..3c138ca58f169 100644 --- a/solution/2300-2399/2391.Minimum Amount of Time to Collect Garbage/README.md +++ b/solution/2300-2399/2391.Minimum Amount of Time to Collect Garbage/README.md @@ -72,7 +72,7 @@ 根据题目描述，每一辆垃圾车从房子 $0$ 出发，收集其中一种垃圾，按顺序前进，直到到达该种垃圾最后出现的房子下标为止。 -因此，我们可以用一个哈希表 $\text{last}$ 记录每种垃圾最后出现的房子下标。我们假设第 $i$ 种垃圾最后一次出现在第 $j$ 个房子，那么第 $i$ 辆车所需要的行驶时间为 $\text{travel}[0] + \text{travel}[1] + \cdots + \text{travel}[j-1]$。这可以通过前缀和来计算。我们累计所有车辆的行驶时间，加上每种垃圾的总收集时间，即可得到答案。 +因此，我们可以用一个哈希表 $\text{last}$ 记录每种垃圾最后出现的房子下标。我们假设第 $i$ 种垃圾最后一次出现在第 $j$ 个房子，那么第 $i$ 辆车所需要的行驶时间为 $\text{travel}[0] + \text{travel}[1] + \cdots + \text{travel}[j-1]$。注意，如果 $j = 0$，则不需要行驶时间。我们通过前缀和累计所有车辆的行驶时间，加上每种垃圾的总收集时间，即可得到答案。 时间复杂度 $O(n)$，空间复杂度 $O(k)$，其中 $n$ 和 $k$ 分别是垃圾的数量和种类。本题中 $k = 3$。 diff --git a/solution/2300-2399/2391.Minimum Amount of Time to Collect Garbage/README_EN.md b/solution/2300-2399/2391.Minimum Amount of Time to Collect Garbage/README_EN.md index 52ffd178dddcd..842b4e1bb0fab 100644 --- a/solution/2300-2399/2391.Minimum Amount of Time to Collect Garbage/README_EN.md +++ b/solution/2300-2399/2391.Minimum Amount of Time to Collect Garbage/README_EN.md @@ -70,7 +70,7 @@ It takes a total of 7 + 15 + 15 = 37 minutes to collect all the garbage. According to the problem description, each garbage truck starts from house $0$, collects one type of garbage, and moves forward in order until it reaches the house index where this type of garbage last appears. -Therefore, we can use a hash table $\text{last}$ to record the house index where each type of garbage last appears. We assume that the $i$-th type of garbage last appears in the $j$-th house, then the driving time required for the $i$-th truck is $\text{travel}[0] + \text{travel}[1] + \cdots + \text{travel}[j-1]$. This can be calculated by prefix sum. We accumulate the driving time of all vehicles, add the total collection time of each type of garbage, and we can get the answer. +Therefore, we can use a hash table $\text{last}$ to record the house index where each type of garbage last appears. We assume that the $i$-th type of garbage last appears in the $j$-th house, then the driving time required for the $i$-th truck is $\text{travel}[0] + \text{travel}[1] + \cdots + \text{travel}[j-1]$. Note, if $j = 0$, no driving time is needed. We accumulate the driving time of all vehicles, add the total collection time of each type of garbage, and we can get the answer. The time complexity is $O(n)$, and the space complexity is $O(k)$, where $n$ and $k$ are the number and types of garbage, respectively. In this problem, $k = 3$. diff --git a/solution/2400-2499/2473.Minimum Cost to Buy Apples/README_EN.md b/solution/2400-2499/2473.Minimum Cost to Buy Apples/README_EN.md index 6d5b0960d7c7c..6442c9e3342d4 100644 --- a/solution/2400-2499/2473.Minimum Cost to Buy Apples/README_EN.md +++ b/solution/2400-2499/2473.Minimum Cost to Buy Apples/README_EN.md @@ -40,7 +40,7 @@
  • 2 <= n <= 1000
    • -
  • 1 <= roads.length <= 1000
    • +
  • 1 <= roads.length <= 2000
  • 1 <= ai, bi <= n
  • ai != bi
  • 1 <= costi <= 105
    • diff --git a/solution/3000-3099/3017.Count the Number of Houses at a Certain Distance II/README.md b/solution/3000-3099/3017.Count the Number of Houses at a Certain Distance II/README.md index b3086bec4840b..efc1789cd92d7 100644 --- a/solution/3000-3099/3017.Count the Number of Houses at a Certain Distance II/README.md +++ b/solution/3000-3099/3017.Count the Number of Houses at a Certain Distance II/README.md @@ -2,7 +2,7 @@ [English Version](/solution/3000-3099/3017.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20II/README_EN.md) - + ## 题目描述 diff --git a/solution/3000-3099/3017.Count the Number of Houses at a Certain Distance II/README_EN.md b/solution/3000-3099/3017.Count the Number of Houses at a Certain Distance II/README_EN.md index aeb741b53c8b2..de672cc5653ea 100644 --- a/solution/3000-3099/3017.Count the Number of Houses at a Certain Distance II/README_EN.md +++ b/solution/3000-3099/3017.Count the Number of Houses at a Certain Distance II/README_EN.md @@ -2,7 +2,7 @@ [中文文档](/solution/3000-3099/3017.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20II/README.md) - + ## Description diff --git a/solution/README.md b/solution/README.md index 9dfe1771d8b49..37d0c65cefb29 100644 --- a/solution/README.md +++ b/solution/README.md @@ -3027,7 +3027,7 @@ | 3014 | [输入单词需要的最少按键次数 I](/solution/3000-3099/3014.Minimum%20Number%20of%20Pushes%20to%20Type%20Word%20I/README.md) | `贪心`,`数学`,`字符串` | 简单 | 第 381 场周赛 | | 3015 | [按距离统计房屋对数目 I](/solution/3000-3099/3015.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20I/README.md) | `广度优先搜索`,`图`,`前缀和` | 中等 | 第 381 场周赛 | | 3016 | [输入单词需要的最少按键次数 II](/solution/3000-3099/3016.Minimum%20Number%20of%20Pushes%20to%20Type%20Word%20II/README.md) | `贪心`,`哈希表`,`字符串`,`计数`,`排序` | 中等 | 第 381 场周赛 | -| 3017 | [按距离统计房屋对数目 II](/solution/3000-3099/3017.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20II/README.md) | `广度优先搜索`,`图`,`前缀和` | 困难 | 第 381 场周赛 | +| 3017 | [按距离统计房屋对数目 II](/solution/3000-3099/3017.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20II/README.md) | `图`,`前缀和` | 困难 | 第 381 场周赛 | | 3018 | [可处理的最大删除操作数 I](/solution/3000-3099/3018.Maximum%20Number%20of%20Removal%20Queries%20That%20Can%20Be%20Processed%20I/README.md) | `数组`,`动态规划` | 困难 | 🔒 | | 3019 | [按键变更的次数](/solution/3000-3099/3019.Number%20of%20Changing%20Keys/README.md) | `字符串` | 简单 | 第 382 场周赛 | | 3020 | [子集中元素的最大数量](/solution/3000-3099/3020.Find%20the%20Maximum%20Number%20of%20Elements%20in%20Subset/README.md) | `数组`,`哈希表`,`枚举` | 中等 | 第 382 场周赛 | diff --git a/solution/README_EN.md b/solution/README_EN.md index c0128d10d54c3..e9d532569f734 100644 --- a/solution/README_EN.md +++ b/solution/README_EN.md @@ -3025,7 +3025,7 @@ Press Control + F(or Command + F on | 3014 | [Minimum Number of Pushes to Type Word I](/solution/3000-3099/3014.Minimum%20Number%20of%20Pushes%20to%20Type%20Word%20I/README_EN.md) | `Greedy`,`Math`,`String` | Easy | Weekly Contest 381 | | 3015 | [Count the Number of Houses at a Certain Distance I](/solution/3000-3099/3015.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20I/README_EN.md) | `Breadth-First Search`,`Graph`,`Prefix Sum` | Medium | Weekly Contest 381 | | 3016 | [Minimum Number of Pushes to Type Word II](/solution/3000-3099/3016.Minimum%20Number%20of%20Pushes%20to%20Type%20Word%20II/README_EN.md) | `Greedy`,`Hash Table`,`String`,`Counting`,`Sorting` | Medium | Weekly Contest 381 | -| 3017 | [Count the Number of Houses at a Certain Distance II](/solution/3000-3099/3017.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20II/README_EN.md) | `Breadth-First Search`,`Graph`,`Prefix Sum` | Hard | Weekly Contest 381 | +| 3017 | [Count the Number of Houses at a Certain Distance II](/solution/3000-3099/3017.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20II/README_EN.md) | `Graph`,`Prefix Sum` | Hard | Weekly Contest 381 | | 3018 | [Maximum Number of Removal Queries That Can Be Processed I](/solution/3000-3099/3018.Maximum%20Number%20of%20Removal%20Queries%20That%20Can%20Be%20Processed%20I/README_EN.md) | `Array`,`Dynamic Programming` | Hard | 🔒 | | 3019 | [Number of Changing Keys](/solution/3000-3099/3019.Number%20of%20Changing%20Keys/README_EN.md) | `String` | Easy | Weekly Contest 382 | | 3020 | [Find the Maximum Number of Elements in Subset](/solution/3000-3099/3020.Find%20the%20Maximum%20Number%20of%20Elements%20in%20Subset/README_EN.md) | `Array`,`Hash Table`,`Enumeration` | Medium | Weekly Contest 382 |