LeetCode-33-搜索旋转排序数组(Search in Rotated Sorted Array)
33. 搜索旋转排序数组
整数数组 nums 按升序排列，数组中的值 互不相同 。
在传递给函数之前，nums 在预先未知的某个下标 k（0 <= k < nums.length）上进行了 旋转，使数组变为 [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]（下标 从 0 开始 计数）。例如， [0,1,2,4,5,6,7] 在下标 3 处经旋转后可能变为 [4,5,6,7,0,1,2] 。
给你 旋转后 的数组 nums 和一个整数 target ，如果 nums 中存在这个目标值 target ，则返回它的下标，否则返回 -1 。
示例 1：
输入：nums = [4,5,6,7,0,1,2], target = 0
输出：4
示例 2：
输入：nums = [4,5,6,7,0,1,2], target = 3
输出：-1
示例 3：
输入：nums = [1], target = 0
输出：-1
提示：
1 <= nums.length <= 5000-10^4 <= nums[i] <= 10^4-
nums中的每个值都 独一无二 - 题目数据保证
nums在预先未知的某个下标上进行了旋转 -10^4 <= target <= 10^4
进阶：你可以设计一个时间复杂度为 O(log n) 的解决方案吗？
- Search in Rotated Sorted Array
Medium
7580665Add to ListShare
There is an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].
Given the array nums after the rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
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
Example 3:
Input: nums = [1], target = 0
Output: -1
Constraints:
1 <= nums.length <= 5000-104 <= nums[i] <= 104- All values of
numsare unique. -
numsis guaranteed to be rotated at some pivot. -104 <= target <= 104
Follow up: Can you achieve this in O(log n) time complexity?
解题思路
L 左指针
R 右指针
M = (L + R) / 2;
只要数组的[L] [M] [R] 不都一样的情况，进行二分查找。
class Solution {
public static int search(int[] arr, int num) {
int L = 0;
int R = arr.length - 1;
int M = 0;
while (L <= R) {
M = (L + R) / 2;
if (arr[M] == num) {
return M;
}
// arr[M] != num
if (arr[L] == arr[M] && arr[M] == arr[R]) {
while (L != M && arr[L] == arr[M]) {
L++;
}
// L和M没撞上，[L]!=[M] L,.....M
if (L == M) {
L = M + 1;
continue;
}
}
// arr[M] != num
// [L] [M] [R] 不都一样的情况
if (arr[L] != arr[M]) { //[L]<[M] L-M左侧一定有序
if (arr[M] > arr[L]) {
if (num >= arr[L] && num < arr[M]) {
R = M - 1;
} else {
L = M + 1;
}
} else { // [L] > [M] M-R 右侧一定有序
if (num > arr[M] && num <= arr[R]) {
L = M + 1;
} else {
R = M - 1;
}
}
} else { // [L] === [M] -> [M]!=[R]
if (arr[M] < arr[R]) {
if (num > arr[M] && num <= arr[R]) {
L = M + 1;
} else {
R = M - 1;
}
} else {
if (num >= arr[L] && num < arr[M]) {
R = M - 1;
} else {
L = M + 1;
}
}
}
}
return -1;
}
}