Problem:

Analysis:

This question can be transferred as the above figure.
two scenarios: which represent by mid1 and mid2.

Code:
public int search(int[] nums, int target) {
if (nums.length == 0 || nums == null) {
return -1;
}
int left = 0, right = nums.length - 1;
while(left <= right) {
int mid = (left + right) / 2;
if (nums[mid] == target) {
return mid;
}else if (nums[mid] < nums[left]) {
if (nums[mid] < target && nums[right] >= target) {
left = mid + 1;
}else {
right = mid -1;
}
}else {
if (nums[left] <= target && nums[mid] > target) {
right = mid - 1;
}else {
left = mid + 1;
}
}
}
return -1;
}
本文详细解析了在旋转排序数组中使用二分查找算法搜索特定元素的策略。针对两种不同场景,介绍了如何调整查找范围,并提供了完整的代码实现。
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