1 题 起初想到了132问题,想着用单调栈去解,后来就干脆暴力了,写了个O(n^2logn)时间复杂度的算法,数据都过了。题目也没给数据范围。结束后对代码改了改,时间复杂度O(n^2)
import java.util.Arrays; import java.util.TreeSet; public class Solution { public static void main(String[] args) { int res = new Solution().TeamNums(new int[]{1,6,3,2,5,4}); System.out.println(res); } /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param height int整型一维数组 舞蹈员身高的一维数组 * @return int整型 */ public int TeamNums (int[] height) { if (height.length < 3) return 0; int ret = work(height); int i = 0, j = height.length - 1; while(i < j) { int tmp = height[i]; height[i] = height[j]; height[j] = tmp; ++i; --j; } ret += work(height); return ret; } private int work(int[] height) { int count = 0; TreeSet<Integer> set = new TreeSet<>(); int[] rank = new int[height.length]; for (int i = height.length - 1; i >= 0; --i) { int idx = Arrays.binarySearch(set.toArray(), height[i]); if (idx < 0) { idx = -(idx + 1); rank[i] = set.size() - idx; // height[i+1..height.length)中有多少个数比height[i]大 } set.add(height[i]); } for (int i = 0; i < height.length; ++i) { for (int k = height.length - 1; k > i; --k) { if (height[k] < height[i]) continue; count += rank[k]; } } return count; } }
2 题 图中链路上节点和的最大值。时间复杂度O(n*m)
public class Solution { public static void main(String[] args) { int res = new Solution().getMaximumResource(new int[][]{{1,0,7},{2,0,6},{3,4,5},{0,3,0},{9,0,20}}); System.out.println(res); } /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param grid int整型二维数组 为n*m 的二维数组 * @return int整型 */ int n, m; public int getMaximumResource (int[][] grid) { if (grid.length == 0 || grid[0].length == 0) return 0; n = grid.length; m = grid[0].length; int maxV = 0; for (int i = 0; i < n; ++i) { for (int j = 0; j < m; ++j) { if (grid[i][j] != 0) { maxV = Math.max(maxV, dfs(grid, i, j, i, j)); } } } return maxV; } static int[][] dir = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; int dfs(int[][] grid, int x, int y, int x0, int y0) { int ret = grid[x][y]; grid[x][y] = 0; int a = 0, b = 0; for (int i = 0; i < 4; ++i) { int xx = x + dir[i][0]; int yy = y + dir[i][1]; if (xx >= 0 && xx < n && yy >= 0 && yy < m && grid[xx][yy] != 0) { int res = dfs(grid, xx, yy, x0, y0); if (res >= a) { b = a; a = res; } else if (res > b) { b = res; } } } return ret + (x == x0 && y == y0 ? a + b: Math.max(a, b)); } }
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