33. Search in Rotated Sorted Array
There is an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= 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 possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
You must write an algorithm with O(log n) runtime complexity.
Example 1:
Example 2:
Example 3:
Constraints:
1 <= nums.length <= 5000-104 <= nums[i] <= 104- All values of
numsare unique. numsis an ascending array that is possibly rotated.-104 <= target <= 104
Solution
class Solution {
public int search(int[] nums, int target) {
int min = findMin(nums);
if (nums[min] == target){
return min;
}
if (target <= nums[nums.length - 1]){
return binarySearch(nums, target, min, nums.length - 1);
}else{
return binarySearch(nums, target, 0, min - 1);
}
}
private int findMin(int[] nums){
int left = 0;
int right = nums.length - 1 - 1;
while(left <= right){
int mid = left + (right - left) / 2;
if (nums[mid] < nums[nums.length - 1]){
right = mid - 1;
}else{
left = mid + 1;
}
}
return left;
}
private int binarySearch(int[] nums, int target, int left, int right){
while(left <= right){
int mid = left + (right - left)/2;
if (nums[mid] == target){
return mid;
}else if (nums[mid] < target){
left = mid + 1;
}else{
right = mid - 1;
}
}
if (nums[left] != target){
return -1;
}else{
return left;
}
}
}
class Solution {
public int search(int[] nums, int target) {
int min = findMin(nums);
if (target > nums[nums.length - 1]){ // target 在第一段
return lowerBound(nums, 0, min -1, target);
}else{
// target 在第二段
return lowerBound(nums, min, nums.length - 1, target);
}
}
// 153
private int findMin(int[] nums){
int left = 0;
int right = nums.length - 1 - 1;
while(left <= right){
int mid = left + (right - left)/2;
if (nums[mid] < nums[nums.length - 1]){
right = mid - 1;
}else{
left = mid + 1;
}
}
return left;
}
// 有序数组中找target的下标
private int lowerBound(int[] nums, int left, int right, int target){
while(left <= right){
int mid = left + (right - left)/2;
if (nums[mid] < target){
left = mid + 1;
}else{
right = mid - 1;
}
}
return nums[left] == target ? left : -1;
}
}
class Solution {
public int search(int[] nums, int target) {
int left = 0;
int right = nums.length - 1;
while (left <= right){ // Case 1: find target
int mid = left + (right - left)/2;
if (nums[mid] == target){
return mid;
}else if (nums[mid] >= nums[left]){ // Case 2: subarray on mid's left is sorted // must have =
if (target >= nums[left] && target < nums[mid]){
right = mid-1;
}else{
left = mid +1;
}
}else{ // Case 3: subarray on mid's right is sorted
if (target <= nums[right] && target > nums[mid]){
left = mid + 1;
}else{
right = mid -1;
}
}
}
return -1;
}
}
// TC: O(logn)
// SC: O(1)
Binary search的题边界要注意
class Solution {
public int search(int[] nums, int target) {
int n = nums.length;
int left = 0;
int right = n - 1;
/*
[4,5,6,7,0,1,2]
l
r
m
*/
while(left <= right){
int mid = left + (right - left)/2;
if (nums[mid] == target){
return mid;
}else if (nums[left] <= nums[mid]){
if (nums[left] <= target && target <= nums[mid]){
right = mid - 1;
}else{
left = mid + 1;
}
}else{
// nums[left] > nums[mid]
if (nums[mid] <= target && target <= nums[right]){
left = mid +1;
}else{
right = mid -1;
}
}
}
return -1;
}
}
// TC: O(logn)
// SC: O(n)
class Solution {
public int search(int[] nums, int target) {
int n = nums.length;
int left = 0;
int right = n-1;
if (n == 1){
if (nums[0] != target){
return -1;
}else{
return 0;
}
}
/* 0 1 2 3 4 5 6
4 5 6 7 0 1 2
t = 3
l
r
m l + (r - l ) /2 = 0 + (6) /2 = 3
if (nums[left] < nums[right]) -> in order sorted
/// not order
if (target < nums[right]) -> right
if (target > nums[left]) -> left part
*/
/*
0 1
[3, 1]
l
r
0 + (1- 0) /2 = 0
m
0 1 2 3 4 5 6 7
[4,5,6,7,8,1,2,3]
l
r
m 0 + (7 - 0) /2 3
*/
while(left <= right){
int mid = left + (right - left)/2;
if (nums[mid] == target){
return mid;
}else{
// move
if (nums[left] < nums[right]){
if (nums[mid] < target){
left = mid + 1;
}else{
right = mid - 1;
}
}else{
// nums[left] > nums[right]
if (target == nums[right]){
return right;
}
if (target == nums[left]){
return left;
}
if (target < nums[right]){
left = left + 1;
}else{
right = right - 1;
}
}
}
}
return -1;
}
}
// TC: O(logn)
// SC: O(1)