来源:力扣(LeetCode) 链接:33. Search in Rotated Sorted Array 解答 解法1 // 时间复杂度O(log n), 空间复杂度O(1) class Solution { public: int searchRotatedSortedArray(vector<int>& nums, int target) { int lo = 0, hi = nums.size(); while (lo != hi) { int mid = lo + (hi -...
For example, Given sorted arrayA = [1,1,1,2,2,3], Your function should return length = 5, and A is now[1,1,2,2,3] 解决方法: // 2015.7.3#include<iostream>usingnamespacestd;intremoveDuplicates(intA[],intn);voidmain(){intA[6]={1,1,1,2,2,3};cout<<"长度是"<<removeDuplicate...
\begin{Code} // LeetCode, Search in Rotated Sorted Array II // 时间复杂度O(n),空间复杂度O(1) class Solution { public: bool search(const vector<int>& nums, int target) { int first = 0, last = nums.size(); while (first != last) { const int mid = first + (last - first)...
\begin{Code} // LeetCode, Search in Rotated Sorted Array II // 时间复杂度O(n),空间复杂度O(1) class Solution { public: bool search(int A[], int n, int target) { int first = 0, last = n; while (first != last) { const int mid = first + (last - first) / 2; if (A[...