题目链接 🔗
解法一(朴素二分查找 无法 AC)
int find(vector<int> &nums, int left, int right, int target) {
while (left < right) {
int mid = left + right >> 1;
if (nums[mid] >= target) {
right = mid;
} else {
left = mid + 1;
}
}
return nums[left] == target ? left : -1;
}
int search(vector<int> &nums, int target) {
int N = nums.size();
int idx = 0;
for (int i = 0; i < N; i++) {
if (nums[i] > nums[i + 1]) {
idx = i;
break;
}
}
int ans = find(nums, 0, idx, target);
if (ans != -1)return ans;
if (idx + 1 < N) ans = find(nums, idx + 1, N - 1, target);
return ans;
}
解法二 二分
题解 🔗
int search2(vector<int> &nums, int target) {
int N = nums.size();
int l = 0, r = N - 1;
while (l < r) {
int mid = l + r + 1 >> 1;
if (nums[mid] >= nums[0]) l = mid;
else r = mid - 1;
}
if (target >= nums[0]) l = 0;
else {
l++;
r = N - 1;
}
while (l < r) {
int mid = l + r >> 1;
if (nums[mid] >= target) r = mid;
else l = mid + 1;
}
return (nums[] == target ? r : -1);
}