Difficulty: Medium
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:
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| Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
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Example 2:
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| Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
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Solution
Language: Java
两次二分
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| class Solution {
public int search(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return -1;
}
int last = nums.length - 1;
int start = 0, end = nums.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (nums[mid] <= nums[last]) {
end = mid;
} else {
start = mid;
}
}
int rot = end;
if (target <= nums[last]) {
return binarySearch(nums, target, start, last);
} else {
return binarySearch(nums, target, 0, rot);
}
}
private int binarySearch(int[] nums, int target, int start, int end) {
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (nums[mid] == target) {
return mid;
} else if (nums[mid] > target) {
end = mid;
} else {
start = mid;
}
}
if (nums[start] == target) {
return start;
}
if (nums[end] == target) {
return end;
}
return -1;
}
}
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一次二分
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| class Solution {
public int search(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return -1;
}
int start = 0, end = nums.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (nums[mid] == target) {
return mid;
}
if (nums[start] < nums[mid]) {
if (nums[start] <= target && target <= nums[mid]) {
end = mid;
} else {
start = mid;
}
} else {
if (nums[mid] <= target && target <= nums[end]) {
start = mid;
} else {
end = mid;
}
}
}
if (nums[start] == target) {
return start;
}
if (nums[end] == target) {
return end;
}
return -1;
}
}
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| class Solution {
public int search(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return -1;
}
int start = 0, end = nums.length - 1;
int last = nums[end];
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (nums[mid] == target) {
return mid;
}
if (nums[mid] <= nums[end]) {
if (nums[mid] <= target && target <= nums[end]) {
start = mid;
} else {
end = mid;
}
} else {
if (target <= nums[end]) {
start = mid;
} else if (target <= nums[mid]) {
end = mid;
} else {
start = mid;
}
}
}
if (nums[start] == target) {
return start;
}
if (nums[end] == target) {
return end;
}
return -1;
}
}
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