Given an array of integers, find if the array contains any duplicates. Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct. 思路1. 对每一个数字都与其后序的数字进行比较,如果相同则返回true,如果不…

准备刷一刷LeetCode了. 题目: ''' Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution. Example: Given nums = [2, 7, 11, 15], target = 9, Becaus…

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order). The replaceme…

Given a collection of numbers that might contain duplicates, return all possible unique permutations. For example,[1,1,2] have the following unique permutations: [ [1,1,2], [1,2,1], [2,1,1] ] 思路1.这题与Permutations的区别在于他允许重复数字,最简单的就是利用1的结果,保存在一个set中,去重复…

Reverse bits of a given 32 bits unsigned integer. For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000). Follow up:If this function…

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority element always exist in the array. 思路.1 排序,选择第n/2个数,调用STL的sort,…

Given an array of integers, find if the array contains any duplicates. Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct. 这个问题比较简单,判断一个数组里的是否有重复出现的数字.代码简单如下: pub…

思路就是从根节点开始向下选节点,依次与sum比较大小,若小,则向下选左右节点其中一个,若大,则接下来判断是否是叶子节点,若是,则返回false 若不是,则上一步选另一节点,再将上述重新执行. 对于叶子节点比较可以: if(root.left == null && root.right == null) { if(root.val == sum) return true; else return false; } 接下来进行递归,看到sum可以每选一层,就将上一层选的节点数值减掉,这样可以计…

对称二叉树,就是左节点的左节点等于右节点的右节点,左节点的右节点等于右节点的左节点. 很自然就想到迭代与递归,可以创建一个新的函数,就是另一个函数不断的判断,返回在主函数. class Solution { public boolean isSymmetric(TreeNode root) { if(root == null) { return true; } else return test(root.left,root.right); } public boolean test(TreeNo…

题目: Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 代码:oj在线测试通过 Runtime: 416 ms # Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next =…