qq_27477039的博客

思考：本题比较简单，是运用广度优先算法，通过队列存储同一深度元素。 public int findBottleLeftValue(TreeNode root){ Queue q = new LinkedList(); q.offer(root); int result = 0; while (!q.isEmpty()) { int size = q.size();

target:Count the number of prime numbers less than a non-negative number n我第一次编写的解决代码，时间复杂度是O（n2）,在40多万时就挂了。 //O(n2),cost too much time public int simplecountPrimes(int n) { int num = 0; for

思考：本题的难点体现在对各种情况能否全部覆盖，已经对细节的把握上。可能出现的各种特殊情况：1.null or empty input 2.+ & - 3.white space 4.max and min of integerpublic class atoi { public int luoAtoi(String str){ if(str==n

@target Write a program to check whether a given number is an ugly number.Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 6, 8 are ugly while 14 is not

Write a program to find the n-th ugly number.思路：一开始，我考虑的是遍历查找，逐个判断是否为丑数，果断耗时太长。//TLE,1600th cost 17.148s public static int nthUglyNumber(int n){ if(n <= 0){ return 0; } int count = 0

There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 提到log(m),自然会想到二分查找。 本题相比一个