给你一个升序排列的整数数组nums，和一个整数 target。假设按照升序排序的数组在预先未知的某个点上进行了旋转.例如，数组[0,1,2,4,5,6,7]可能变为[4,5,6,7,0,1,2]。请你在数组中搜索 target ，如果数组中存在这个目标值，则返回它的索引，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解法:
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75object RotateSearch {
def main(args: Array[String]): Unit = {
println(search(Array(4, 5, 6, 7, 0, 1, 2), -1))
println(search(Array(1, 2), 2))
println(search(Array(1), 0))
println(search(Array(3, 1), 0))
}
def search(nums: Array[Int], target: Int): Int = {
var l = 0
var r = nums.length - 1
if (target > nums(r) && target < nums(l)) return -1
else if (nums(l) == target) return l
else if (nums(r) == target) return r
if (nums.length == 1) {
if (nums(0) == target) return 0 else return -1
}
while (l < r) {
val m = (l + r) / 2
if (nums(m) == target) {
return m
}
if (nums(l) == target) {
return l
}
if (nums(r) == target) {
return r
}
if (nums(l) > nums(r)) {
if (nums(m) > nums(l)) {
if (nums(m) > target) {
if (nums(l) > target) {
l = m + 1
} else if (nums(l) < target) {
r = m - 1
}
} else if (nums(m) < target) {
l = m + 1
}
}
if (nums(m) < nums(r)) {
if (nums(m) > target) {
r = m - 1
} else if (nums(m) < target) {
if (nums(r) < target) {
r = m - 1
} else if (nums(r) > target) {
l = m + 1
}
}
}
} else if (nums(l) < nums(r)) {
if (nums(m) < target) {
l = m + 1
} else if (nums(m) > target) {
r = m - 1
}
}
if (l == m) {
l = m + 1
}
if (r == m) {
r = m - 1
}
}
-1
}
}