题目: 题解:
class Solution {
public:double quickMul(double x, long long N) {if (N 0) {return 1.0;}double y quickMul(x, N / 2);return N % 2 0 ? y * y : y * y * x;}double myPow(double x, int n) {long long N n;return N > 0 ? qu…
解题思路: class Solution {public boolean canFinish(int numCourses, int[][] prerequisites) {int[] inDegree new int[numCourses];//存每个结点的入度List<List<Integer>> res new ArrayList<>();//存结点之间依赖关系Queue<Integer>…