标签：const 结果 get put duplicate return code ios before
题目描述：
Suppose an array sorted in ascending order 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 return -1. You may assume no duplicate exists in the array. Your algorithm‘s runtime complexity must be in the order of O(log n). Example 1: Input: nums = [4,5,6,7,0,1,2], target = 0 Output: 4 Example 2: Input: nums = [4,5,6,7,0,1,2], target = 3 Output: -1
解题思路：由于数组中不存在重复的元素，题目的复杂性就有所降低。为了保持思路的简洁，我们只关注数组三个位置的值：*first, *mid, *last。
第一步：判断下列情况哪一种成立：(1) *first <= *mid；(2) *first > *mid。
第二步：if (1)，说明first和mid之间是单调递增的，可以判断target是否在first和mid之间，若在，则令last=mid，若不在，则令first=mid+1，再从第一步重新开始；if (2)，说明mid和last之间是单调递增的，可以判断target是否在mid和last之间，若在，则令first=mid+1，若不在，则令last=mid，再从第一步重新开始。
重复执行第一、二步，直至*mid==target或first==last+1。这道题求解思路的关键在找到单调递增区域，然后迭代，直至满足终止条件。
注意这里的终止条件，一开始的*first、*last分别对应着数组的首个值和末尾值。为了让mid能够迭代到数组的末尾位置，即mid==last，根据求中点的表达式mid=first+(last-first)/2，可以解出first==last。当target位于数组末尾位置时，为了能返回target的位置，循环的终止条件应设为first==last+1。
参考代码：
#include <vector> #include <iostream> using namespace std; class Solution { public: int search(const vector<int>& nums, int target) { int first = 0, last = nums.size() - 1; while (first != last + 1) { const int mid = first + (last - first) / 2; if (nums[mid] == target) return mid; if (nums[first] <= nums[mid]) { if (nums[first] <= target && target < nums[mid]) last = mid; else first = mid + 1; } else { if (nums[mid] < target && target <= nums[last]) first = mid + 1; else last = mid; } } return -1; } }; int main() { int a[] = {2, 5, 6, 0, 1}; Solution solu; vector<int> vec_arr(a, a+5); int index = solu.search(vec_arr, 1); cout << "target index: " << index << endl; return 0; }
运行结果为：
target index: 4
33. Search in Rotated Sorted Array
标签：const 结果 get put duplicate return code ios before
原文地址：https://www.cnblogs.com/pursuiting/p/10427938.html
