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题目大意:有A个0和B个1,每次取两个出来进行{XNOR,NAND,NOR}操作生成一个新的0/1,直到只剩一个元素.问最后是否可能剩下一个0,是否可能剩下一个1. XNOR 比较特殊 a XNOR b = a xor b xor 1,所以可以发现答案只跟B的奇偶性有关 NAND和NOR 找规律发现,当A和B很大的时候,答案肯定是0和1都可以,所以手动或者用程序算出A和B比较小的情况 (A+B>3的答案都是B) 当时最后做的这道题,,有点烧脑 被前面的题搞的有点疲惫 最后没做出来 其实难度不大…
Remove Duplicates from Sorted Array II Follow up for "Remove Duplicates":What if duplicates are allowed at most twice? For example,Given sorted array A = [1,1,1,2,2,3], Your function should return length = 5, and A is now [1,1,2,2,3].     用两个指针,…
Search in Rotated Sorted Array II Follow up for "Search in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the array.   与I类似,只是在二…
Remove Duplicates from Sorted Array Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this in place with constant memory.…
Search in Rotated Sorted Array Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target value to search. If found in the array return its index, otherwise retu…
Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). Find the minimum element. The array may contain duplicates. 解题思路: 参考Java for LeetCode 081 Search in Rotated Sorted Array II J…
LintCode 160. Find Minimum in Rotated Sorted Array II (Medium) LeetCode 154. Find Minimum in Rotated Sorted Array II (Hard) 解法1 自己做了半个小时, 想分情况判断num[mid]与num[start], num[mid]与num[end]的大小关系, 每个关系分>, =, <的情况, 于是就是9种情况... >, >, L = M + 1 >, =,…
以下三个问题的典型的两个指针处理数组的问题,一个指针用于遍历,一个指针用于指向当前处理到位置 一:Remove Element Given an array and a value, remove all instances of that value in place and return the new length. The order of elements can be changed. It doesn't matter what you leave beyond the new l…
LeetCode 81 Search in Rotated Sorted Array II [binary search] <c++> 给出排序好的一维有重复元素的数组,随机取一个位置断开,把前半部分接到后半部分后面,得到一个新数组,在新数组中查找给定数是否存在,时间复杂度限制\(O(log_2n)\) C++ 因为有重复元素存在,nums[l] <= nums[mid]不能说明[l,mid]区间内一定是单调的,比如数组[1,2,3,1,1,1,1],但是严格小于和严格大于的情况还是可以…
LeetCode 80 Remove Duplicates from Sorted Array II [Array/auto] <c++> 给出排序好的一维数组,如果一个元素重复出现的次数大于两次,删除多余的复制,返回删除后数组长度,要求不另开内存空间. C++ 献上自己丑陋无比的代码.相当于自己实现一个带计数器的unique函数 class Solution { public: int removeDuplicates(std::vector<int>& nums) {…