From 23290065ba0c3b93a7c9f18f05e12a622e58dd27 Mon Sep 17 00:00:00 2001 From: algorithmzuo Date: Thu, 2 Mar 2023 18:30:34 +0800 Subject: [PATCH] modify code --- .../第03期/mca_04/Code07_LIS.java | 32 ++++ .../mca_04/Code07_SwimInRisingWater.java | 155 ------------------ 2 files changed, 32 insertions(+), 155 deletions(-) create mode 100644 MCA算法突击课/第03期/mca_04/Code07_LIS.java delete mode 100644 MCA算法突击课/第03期/mca_04/Code07_SwimInRisingWater.java diff --git a/MCA算法突击课/第03期/mca_04/Code07_LIS.java b/MCA算法突击课/第03期/mca_04/Code07_LIS.java new file mode 100644 index 0000000..79785a1 --- /dev/null +++ b/MCA算法突击课/第03期/mca_04/Code07_LIS.java @@ -0,0 +1,32 @@ +package 第03期.mca_04; + +// 本题测试链接 : https://leetcode.cn/problems/longest-increasing-subsequence +public class Code07_LIS { + + public static int lengthOfLIS(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int[] ends = new int[arr.length]; + ends[0] = arr[0]; + int right = 0; + int max = 1; + for (int i = 1; i < arr.length; i++) { + int l = 0; + int r = right; + while (l <= r) { + int m = (l + r) / 2; + if (arr[i] > ends[m]) { + l = m + 1; + } else { + r = m - 1; + } + } + right = Math.max(right, l); + ends[l] = arr[i]; + max = Math.max(max, l + 1); + } + return max; + } + +} \ No newline at end of file diff --git a/MCA算法突击课/第03期/mca_04/Code07_SwimInRisingWater.java b/MCA算法突击课/第03期/mca_04/Code07_SwimInRisingWater.java deleted file mode 100644 index 932160b..0000000 --- a/MCA算法突击课/第03期/mca_04/Code07_SwimInRisingWater.java +++ /dev/null @@ -1,155 +0,0 @@ -package 第03期.mca_04; - -import java.util.Arrays; -import java.util.PriorityQueue; - -// 在一个 n x n 的整数矩阵 grid 中， -// 每一个方格的值 grid[i][j] 表示位置 (i, j) 的平台高度。 -// 当开始下雨时，在时间为 t 时，水池中的水位为 t 。 -// 你可以从一个平台游向四周相邻的任意一个平台，但是前提是此时水位必须同时淹没这两个平台。 -// 假定你可以瞬间移动无限距离，也就是默认在方格内部游动是不耗时的。 -// 当然，在你游泳的时候你必须待在坐标方格里面。 -// 你从坐标方格的左上平台 (0，0) 出发。 -// 返回 你到达坐标方格的右下平台 (n-1, n-1) 所需的最少时间 。 -// 测试链接 ：https://leetcode.cn/problems/swim-in-rising-water -public class Code07_SwimInRisingWater { - - // 并查集的解法 - public static int swimInWater1(int[][] grid) { - // 行号 - int n = grid.length; - // 列号 - int m = grid[0].length; - // [0,0,5] - // [0,1,3].... - int[][] points = new int[n * m][3]; - int pi = 0; - for (int i = 0; i < n; i++) { - for (int j = 0; j < m; j++) { - points[pi][0] = i; - points[pi][1] = j; - points[pi++][2] = grid[i][j]; - } - } - // 所有格子小对象，生成好了! - // 排序！[a,b,c] [d,e,f] - Arrays.sort(points, (a, b) -> a[2] - b[2]); - // 生成并查集！n * m - // 初始化的时候，把所有格子独自成一个集合！ - UnionFind uf = new UnionFind(n, m); - int ans = 0; - for (int i = 0; i < points.length; i++) { - int r = points[i][0]; - int c = points[i][1]; - int v = points[i][2]; - if (r > 0 && grid[r - 1][c] <= v) { - uf.union(r, c, r - 1, c); - } - if (r < n - 1 && grid[r + 1][c] <= v) { - uf.union(r, c, r + 1, c); - } - if (c > 0 && grid[r][c - 1] <= v) { - uf.union(r, c, r, c - 1); - } - if (c < m - 1 && grid[r][c + 1] <= v) { - uf.union(r, c, r, c + 1); - } - if (uf.isSameSet(0, 0, n - 1, m - 1)) { - ans = v; - break; - } - } - return ans; - } - - public static class UnionFind { - public int col; - public int pointsSize; - public int[] father; - public int[] size; - public int[] help; - - public UnionFind(int n, int m) { - col = m; - pointsSize = n * m; - father = new int[pointsSize]; - size = new int[pointsSize]; - help = new int[pointsSize]; - for (int i = 0; i < pointsSize; i++) { - father[i] = i; - size[i] = 1; - } - } - - private int find(int i) { - int hi = 0; - while (i != father[i]) { - help[hi++] = i; - i = father[i]; - } - while (hi > 0) { - father[help[--hi]] = i; - } - return i; - } - - private int index(int i, int j) { - return i * col + j; - } - - public void union(int row1, int col1, int row2, int col2) { - int f1 = find(index(row1, col1)); - int f2 = find(index(row2, col2)); - if (f1 != f2) { - if (size[f1] >= size[f2]) { - father[f2] = f1; - size[f1] += size[f2]; - } else { - father[f1] = f2; - size[f2] += size[f1]; - } - } - } - - public boolean isSameSet(int row1, int col1, int row2, int col2) { - return find(index(row1, col1)) == find(index(row2, col2)); - } - - } - - // Dijkstra算法 - public static int swimInWater2(int[][] grid) { - int n = grid.length; - int m = grid[0].length; - PriorityQueue heap = new PriorityQueue<>((a, b) -> a[2] - b[2]); - boolean[][] visited = new boolean[n][m]; - heap.add(new int[] { 0, 0, grid[0][0] }); - int ans = 0; - while (!heap.isEmpty()) { - int r = heap.peek()[0]; - int c = heap.peek()[1]; - int v = heap.peek()[2]; - heap.poll(); - if (visited[r][c]) { - continue; - } - visited[r][c] = true; - if (r == n - 1 && c == m - 1) { - ans = v; - break; - } - add(grid, heap, visited, r - 1, c, v); - add(grid, heap, visited, r + 1, c, v); - add(grid, heap, visited, r, c - 1, v); - add(grid, heap, visited, r, c + 1, v); - } - return ans; - } - - public static void add(int[][] grid, PriorityQueue heap, boolean[][] visited, int r, int c, int preV) { - if (r >= 0 && r < grid.length && c >= 0 && c < grid[0].length && !visited[r][c]) { - heap.add(new int[] { r, c, preV + Math.max(0, grid[r][c] - preV) }); - } - } - -}