Leetcode70场双周赛-第三题2146. 价格范围内最高排名的 K 样物品
作者：互联网
https://leetcode-cn.com/problems/k-highest-ranked-items-within-a-price-range/题目描述
解题思路
核心办法是：BFS+优先队列
第一个思想，就是从起点开始，向四周进行BFS。需要注意的是记录是否访问viewed的数组。
第二个思想，排序，按照BFS的方式遍历之后，先遍历到的层级的点，一定是距离最小的。
那么，就要考虑相同层级的点之间如何排序。这个排序的规则已经由题目规定，按照价格、坐标来排序。排序的关键就是写好优先队列的排序方法。
解题代码
方法一
import java.util.*;
public class Solution5973 {
public static void main(String[] args) {
int[][] grid = {{1, 2, 0, 1}, {1, 3, 3, 1}, {0, 2, 5, 1}};
int[] pricing = {2, 3};
int[] start = {2, 3};
int k = 2;
List<List<Integer>> res = new Solution5973().highestRankedKItems(grid, pricing, start, k);
System.out.println(res);
}
public List<List<Integer>> highestRankedKItems(int[][] grid, int[] pricing, int[] start, int k) {
Comparator<int[]> c = new Comparator<int[]>() {
@Override
public int compare(int[] o1, int[] o2) {
int o1x = o1[0];
int o1y = o1[1];
int o2x = o2[0];
int o2y = o2[1];
if (grid[o1x][o1y] < grid[o2x][o2y]) {
return -1;
} else if (grid[o1x][o1y] > grid[o2x][o2y]) {
return 1;
} else {
if (o1x < o2x) {
return -1;
} else if (o1x > o2x) {
return 1;
} else {
return o1y - o2y;
}
}
}
};
List<List<Integer>> res = new ArrayList<>();
int grid_x = grid.length;
int grid_y = grid[0].length;
int low = pricing[0];
int high = pricing[1];
// 定义BFS的观察路径
boolean[][] viewed = new boolean[grid_x][grid_y];
// 定义queue
LinkedList<int[]> queue = new LinkedList<>();
queue.add(start);
while (!queue.isEmpty()) {
int size = queue.size();
PriorityQueue<int[]> priorityQueue = new PriorityQueue<>(c);
for (int i = 0; i < size; i++) {
int[] s = queue.poll();
int x = s[0];
int y = s[1];
//处理当前点
if (x >= 0 && x <= grid_x && y >= 0 && y <= grid_y &&
!viewed[x][y] && grid[x][y] != 0) {
queue.offer(new int[]{x, y});
viewed[x][y] = true;
if (grid[x][y] >= low && grid[x][y] <= high) {
priorityQueue.add(new int[]{x, y});
}
// 扩展当前点的周围4个点
if (x - 1 >= 0 && !viewed[x - 1][y] && grid[x - 1][y] != 0) {
queue.offer(new int[]{x - 1, y});
}
if (x + 1 < grid_x && !viewed[x + 1][y] && grid[x + 1][y] != 0) {
queue.offer(new int[]{x + 1, y});
}
if (y - 1 >= 0 && !viewed[x][y - 1] && grid[x][y - 1] != 0) {
queue.offer(new int[]{x, y - 1});
}
if (y + 1 < grid_y && !viewed[x][y + 1] && grid[x][y + 1] != 0) {
queue.offer(new int[]{x, y + 1});
}
}
}
//从优先队列中取数字
while (!priorityQueue.isEmpty()) {
int[] r = priorityQueue.poll();
ArrayList<Integer> arrayList = new ArrayList<>();
arrayList.add(r[0]);
arrayList.add(r[1]);
res.add(arrayList);
if (res.size() == k) {
return res;
}
}
}
return res;
}
}
方法二
public class Solution5973_2 {
public static void main(String[] args) {
int[][] grid = {{1, 2, 0, 1}, {1, 3, 3, 1}, {0, 2, 5, 1}};
int[] pricing = {2, 3};
int[] start = {2, 3};
int k = 2;
List<List<Integer>> res = new Solution5973_2().highestRankedKItems(grid, pricing, start, k);
System.out.println(res);
}
public List<List<Integer>> highestRankedKItems(int[][] grid, int[] pricing, int[] start, int k) {
Comparator<int[]> c = new Comparator<int[]>() {
@Override
public int compare(int[] o1, int[] o2) {
int o1x = o1[0];
int o1y = o1[1];
int o2x = o2[0];
int o2y = o2[1];
if (grid[o1x][o1y] < grid[o2x][o2y]) {
return -1;
} else if (grid[o1x][o1y] > grid[o2x][o2y]) {
return 1;
} else {
if (o1x < o2x) {
return -1;
} else if (o1x > o2x) {
return 1;
} else {
return o1y - o2y;
}
}
}
};
List<List<Integer>> res = new ArrayList<>();
int grid_x = grid.length;
int grid_y = grid[0].length;
LinkedList<int[]> queue = new LinkedList<>();
boolean[][] viewed = new boolean[grid_x][grid_y];
viewed[start[0]][start[1]] = true;
if (grid[start[0]][start[1]] != 0) {
queue.add(start);
}
if (grid[start[0]][start[1]] <= pricing[1] && grid[start[0]][start[1]] >= pricing[0]) {
ArrayList<Integer> arrayList = new ArrayList<>();
arrayList.add(start[0]);
arrayList.add(start[1]);
res.add(arrayList);
if (res.size() == k) {
return res;
}
}
int low = pricing[0];
int high = pricing[1];
while (!queue.isEmpty()) {
int size = queue.size();
PriorityQueue<int[]> priorityQueue = new PriorityQueue<>(c);
for (int i = 0; i < size; i++) {
int[] s = queue.poll();
int x = s[0];
int y = s[1];
// 扩展4个点
if (x - 1 >= 0 && !viewed[x - 1][y] && grid[x - 1][y] != 0) {
queue.offer(new int[]{x - 1, y});
viewed[x - 1][y] = true;
if (grid[x - 1][y] >= low && grid[x - 1][y] <= high) {
priorityQueue.add(new int[]{x - 1, y});
}
}
if (x + 1 < grid_x && !viewed[x + 1][y] && grid[x + 1][y] != 0) {
queue.offer(new int[]{x + 1, y});
viewed[x + 1][y] = true;
if (grid[x + 1][y] >= low && grid[x + 1][y] <= high) {
priorityQueue.add(new int[]{x + 1, y});
}
}
if (y - 1 >= 0 && !viewed[x][y - 1] && grid[x][y - 1] != 0) {
queue.offer(new int[]{x, y - 1});
viewed[x][y - 1] = true;
if (grid[x][y - 1] >= low && grid[x][y - 1] <= high) {
priorityQueue.add(new int[]{x, y - 1});
}
}
if (y + 1 < grid_y && !viewed[x][y + 1] && grid[x][y + 1] != 0) {
queue.offer(new int[]{x, y + 1});
viewed[x][y + 1] = true;
if (grid[x][y + 1] >= low && grid[x][y + 1] <= high) {
priorityQueue.add(new int[]{x, y + 1});
}
}
}
//从优先队列中取数字
while (!priorityQueue.isEmpty()) {
int[] r = priorityQueue.poll();
ArrayList<Integer> arrayList = new ArrayList<>();
arrayList.add(r[0]);
arrayList.add(r[1]);
res.add(arrayList);
if (res.size() == k) {
return res;
}
}
}
return res;
}
}
两个方法的区别就是在判断点的时机
方法一是在点的出队的时候，方法二是在点入队之前。
解题结果
标签：Leetcode70,int,res,双周,queue,2146,grid,&&,new 来源： https://blog.csdn.net/Kangyucheng/article/details/122690542