Merge pull request #615 from LetMeFly666/3242

添加问题“3242.设计相邻元素求和服务”的代码和题解

Author: LetMeFly666

Codes/3242-design-neighbor-sum-service.cpp

@@ -0,0 +1,53 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2024-11-09 09:55:46
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2024-11-09 10:06:29
+ */
+#ifdef _WIN32
+#include "_[1,2]toVector.h"
+#endif
+
+const int adj[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
+const int dia[4][2] = {{-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
+
+class NeighborSum {
+private:
+    vector<pair<int, int>> cache;
+public:
+    NeighborSum(vector<vector<int>>& grid) {
+        int n = grid.size();
+        cache.resize(n * n);
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < n; j++) {
+                int cntAdj = 0, cntDia = 0;
+                for (int k = 0; k < 4; k++) {
+                    int x = i + adj[k][0], y = j + adj[k][1];
+                    if (x >= 0 && x < n && y >= 0 && y < n) {
+                        cntAdj += grid[x][y];
+                    }
+                    x = i + dia[k][0], y = j + dia[k][1];
+                    if (x >= 0 && x < n && y >= 0 && y < n) {
+                        cntDia += grid[x][y];
+                    }
+                }
+                cache[grid[i][j]] = {cntAdj, cntDia};
+            }
+        }
+    }
+    
+    int adjacentSum(int value) {
+        return cache[value].first;
+    }
+    
+    int diagonalSum(int value) {
+        return cache[value].second;
+    }
+};
+
+/**
+ * Your NeighborSum object will be instantiated and called as such:
+ * NeighborSum* obj = new NeighborSum(grid);
+ * int param_1 = obj->adjacentSum(value);
+ * int param_2 = obj->diagonalSum(value);
+ */

Codes/3242-design-neighbor-sum-service.go

@@ -0,0 +1,50 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2024-11-09 10:20:14
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2024-11-09 10:42:32
+ */
+package main
+
+var direction = []struct{x, y int}{{-1, 0}, {1, 0}, {0, -1}, {0, 1}, {-1, -1}, {-1, 1}, {1, -1}, {1, 1}}
+type Value [][2]int
+
+type NeighborSum struct {
+    cache Value
+}
+
+
+func Constructor(grid [][]int) NeighborSum {
+    n := len(grid)
+    var neighborSum NeighborSum
+    neighborSum.cache = make(Value, n * n)
+    for i, row := range grid {
+        for j, v := range row {
+            for k, d := range direction {
+                x, y := i + d.x, j + d.y
+                if x >= 0 && x < n && y >= 0 && y < n {
+                    neighborSum.cache[v][k / 4] += grid[x][y]
+                }
+            }
+        }
+    }
+    return neighborSum
+}
+
+
+func (this *NeighborSum) AdjacentSum(value int) int {
+    return this.cache[value][0]
+}
+
+
+func (this *NeighborSum) DiagonalSum(value int) int {
+    return this.cache[value][1]
+}
+
+
+/**
+ * Your NeighborSum object will be instantiated and called as such:
+ * obj := Constructor(grid);
+ * param_1 := obj.AdjacentSum(value);
+ * param_2 := obj.DiagonalSum(value);
+ */

Codes/3242-design-neighbor-sum-service.java

@@ -0,0 +1,40 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2024-11-09 10:15:35
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2024-11-09 10:19:36
+ */
+class NeighborSum {
+    private static final int[][] direction = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}, {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
+    private int[][] cache;
+
+    public NeighborSum(int[][] grid) {
+        int n = grid.length;
+        cache = new int[n * n][2];
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < n; j++) {
+                for (int k = 0; k < 8; k++) {
+                    int x = i + direction[k][0], y = j + direction[k][1];
+                    if (x >= 0 && x < n && y >= 0 && y < n) {
+                        cache[grid[i][j]][k / 4] += grid[x][y];
+                    }
+                }
+            }
+        }
+    }
+    
+    public int adjacentSum(int value) {
+        return cache[value][0];
+    }
+    
+    public int diagonalSum(int value) {
+        return cache[value][1];
+    }
+}
+
+/**
+ * Your NeighborSum object will be instantiated and called as such:
+ * NeighborSum obj = new NeighborSum(grid);
+ * int param_1 = obj.adjacentSum(value);
+ * int param_2 = obj.diagonalSum(value);
+ */

Codes/3242-design-neighbor-sum-service.py

@@ -0,0 +1,31 @@
+'''
+Author: LetMeFly
+Date: 2024-11-09 10:07:15
+LastEditors: LetMeFly.xyz
+LastEditTime: 2024-11-09 10:15:02
+'''
+from typing import List
+
+direction = [[-1, 0], [1, 0], [0, -1], [0, 1], [-1, -1], [1, 1], [-1, 1], [1, -1]]
+
+class NeighborSum:
+    def __init__(self, grid: List[List[int]]):
+        n = len(grid)
+        self.cache = [[0, 0] for _ in range(n * n)]
+        for i in range(n):
+            for j in range(n):
+                for th, (x, y) in enumerate(direction):
+                    if 0 <= x + i < n and 0 <= y + j < n:
+                        self.cache[grid[i][j]][th // 4] += grid[x + i][y + j]
+
+    def adjacentSum(self, value: int) -> int:
+        return self.cache[value][0]
+
+    def diagonalSum(self, value: int) -> int:
+        return self.cache[value][1]
+
+
+# Your NeighborSum object will be instantiated and called as such:
+# obj = NeighborSum(grid)
+# param_1 = obj.adjacentSum(value)
+# param_2 = obj.diagonalSum(value)

README.md

@@ -718,6 +718,7 @@
 |3216.交换后字典序最小的字符串|简单|<a href="https://leetcode.cn/problems/lexicographically-smallest-string-after-a-swap/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/10/30/LeetCode%203216.%E4%BA%A4%E6%8D%A2%E5%90%8E%E5%AD%97%E5%85%B8%E5%BA%8F%E6%9C%80%E5%B0%8F%E7%9A%84%E5%AD%97%E7%AC%A6%E4%B8%B2/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/143362223" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/lexicographically-smallest-string-after-a-swap/solutions/2971048/letmefly-3216jiao-huan-hou-zi-dian-xu-zu-nxp9/" target="_blank">LeetCode题解</a>|
 |3222.求出硬币游戏的赢家|简单|<a href="https://leetcode.cn/problems/find-the-winning-player-in-coin-game/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/11/05/LeetCode%203222.%E6%B1%82%E5%87%BA%E7%A1%AC%E5%B8%81%E6%B8%B8%E6%88%8F%E7%9A%84%E8%B5%A2%E5%AE%B6/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/143501415" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-the-winning-player-in-coin-game/solutions/2977629/letmefly-3222qiu-chu-ying-bi-you-xi-de-y-08j3/" target="_blank">LeetCode题解</a>|
 |3226.使两个整数相等的位更改次数|简单|<a href="https://leetcode.cn/problems/number-of-bit-changes-to-make-two-integers-equal/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/11/02/LeetCode%203226.%E4%BD%BF%E4%B8%A4%E4%B8%AA%E6%95%B4%E6%95%B0%E7%9B%B8%E7%AD%89%E7%9A%84%E4%BD%8D%E6%9B%B4%E6%94%B9%E6%AC%A1%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/143448117" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/number-of-bit-changes-to-make-two-integers-equal/solutions/2974558/letmefly-3226shi-liang-ge-zheng-shu-xian-j01h/" target="_blank">LeetCode题解</a>|
+|3242.设计相邻元素求和服务|简单|<a href="https://leetcode.cn/problems/design-neighbor-sum-service/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/11/11/LeetCode%203242.%E8%AE%BE%E8%AE%A1%E7%9B%B8%E9%82%BB%E5%85%83%E7%B4%A0%E6%B1%82%E5%92%8C%E6%9C%8D%E5%8A%A1/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/143698347" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/design-neighbor-sum-service/solutions/2985408/letmefly-3242she-ji-xiang-lin-yuan-su-qi-mc7m/" target="_blank">LeetCode题解</a>|
 |3254.长度为K的子数组的能量值I|中等|<a href="https://leetcode.cn/problems/find-the-power-of-k-size-subarrays-i/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/11/06/LeetCode%203254.%E9%95%BF%E5%BA%A6%E4%B8%BAK%E7%9A%84%E5%AD%90%E6%95%B0%E7%BB%84%E7%9A%84%E8%83%BD%E9%87%8F%E5%80%BCI/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/143575677" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-the-power-of-k-size-subarrays-i/solutions/2979562/letmefly-3254chang-du-wei-k-de-zi-shu-zu-iu8u/" target="_blank">LeetCode题解</a>|
 |3255.长度为K的子数组的能量值II|中等|<a href="https://leetcode.cn/problems/find-the-power-of-k-size-subarrays-ii/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/11/07/LeetCode%203255.%E9%95%BF%E5%BA%A6%E4%B8%BAK%E7%9A%84%E5%AD%90%E6%95%B0%E7%BB%84%E7%9A%84%E8%83%BD%E9%87%8F%E5%80%BCII/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/143591327" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-the-power-of-k-size-subarrays-ii/solutions/2980432/letmefly-3255chang-du-wei-k-de-zi-shu-zu-rags/" target="_blank">LeetCode题解</a>|
 |3259.超级饮料的最大强化能量|中等|<a href="https://leetcode.cn/problems/maximum-energy-boost-from-two-drinks/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/11/01/LeetCode%203259.%E8%B6%85%E7%BA%A7%E9%A5%AE%E6%96%99%E7%9A%84%E6%9C%80%E5%A4%A7%E5%BC%BA%E5%8C%96%E8%83%BD%E9%87%8F/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/143429899" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-energy-boost-from-two-drinks/solutions/2973620/letmefly-3259chao-ji-yin-liao-de-zui-da-eeusa/" target="_blank">LeetCode题解</a>|