modify code

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;
+	}
+
+}

MCA算法突击课/第03期/mca_04/Code07_SwimInRisingWater.java

@@ -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<int[]> 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<int[]> 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) });
-		}
-	}
-
-}