33. 搜索旋转排序数组（C++题解含VS可运行源程序）
1.题解
- 先利用二分法思想找出旋转下标；
- 再利用二分法找出目标值。
解释：
- 只要是要求时间复杂度在log（n）的查找，一般你就用二分查找上面靠肯定没错啦，但是二分查找的数组必须是有序的；
- 首先利用二分查找的思想找到旋转数组的旋转位置，这样这个位置两边的子数组就都是有序的了，
- 然后根据这个位置的值得情况判断是在左边进行二分还是右边进行二分查找。
2.力扣C++源码
class Solution {
public:
int search(vector<int>& nums, int target) {
int left = 0;
int right = nums.size() - 1;
int mid = 0;
if (nums[left] <= nums[right]) {
return binsearch(nums, target, left, right);
}
while (left < right) {
mid = (left + right) / 2;
if (nums[mid] > nums[left]) {
left = mid;
}
else if (nums[mid] < nums[right]) {
right = mid;
}
if (right - left <= 1) {
break;
}
}
if (target >= nums[0]) {
return binsearch(nums, target, 0, left);
}
else {
return binsearch(nums, target, right, nums.size() - 1);
}
}
int binsearch(vector<int>& nums, int target, int left, int right) {
int mid = 0;
while (left <= right) {
mid = (left + right) / 2;
if (nums[mid] < target) {
left = mid + 1;
}
else if (nums[mid] > target) {
right = mid - 1;
}
else {
return mid;
}
}
return -1;
}
};
3.VS可运行源程序
#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<limits>
#include<algorithm>
#include<math.h>
#pragma warning(disable:4996)
using namespace std;
//首先利用二分查找的思想找到旋转数组的旋转位置，这样这个位置两边的子数组就都是有序的了，
//然后根据这个位置的值得情况判断是在左边进行二分还是右边进行二分查找。
class Solution {
public:
int search(vector<int>& nums, int target) {
int left = 0;
int right = nums.size() - 1;
int mid = 0;
//数组升序，直接调用二分法返回结果
if (nums[left] <= nums[right]) {
return binsearch(nums, target, left, right);
}
//利用二分法思想找出旋转下标
while (left < right) {
mid = (left + right) / 2;
if (nums[mid] > nums[left]) {
left = mid;
}
else if (nums[mid] < nums[right]) {
right = mid;
}
if (right - left <= 1) {
break;
}
}//right为旋转下标，left为旋转下标的前一个值(数组最大值)下标
//目标值在前半部分
if (target >= nums[0]) {
return binsearch(nums, target, 0, left);
}
//目标值在后半部分
else {
return binsearch(nums, target, right, nums.size() - 1);
}
}
//二分法查找
int binsearch(vector<int>& nums, int target, int left, int right) {
int mid = 0;
while (left <= right) {
mid = (left + right) / 2;
if (nums[mid] < target) {
left = mid + 1;
}
else if (nums[mid] > target) {
right = mid - 1;
}
else {
return mid;
}
}
return -1;
}
};
int main()
{
printf("输入数组大小：");
int n;
scanf("%d", &n);
printf("输入数组元素：");
vector<int> nums;
int num;
for (int i = 0; i < n; i++) {
scanf("%d", &num);
nums.push_back(num);
}
printf("输入目标值：");
int target;
scanf("%d", &target);
Solution test;
int res = test.search(nums, target);
printf("目标值的下标为：%d", res);
printf("\n");
system("pause");
return 0;
}