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原题
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:
Input: nums = [4,5,6,7,0,1,2]
, target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2]
, target = 3
Output: -1
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代码
class Solution {
public:
int findP(vector<int>& nums) {
if (nums[0] < nums[nums.size() - 1]) return -1;
int l = 0, h = nums.size() - 1, m = 0;
while (l < h) {
m = l + (h - l + 1) / 2;
if (nums[m] < nums[0]) {
h = m - 1;
// h = m;
} else {
l = m;
// l = m + 1;
}
}
return l;
}
int findT(vector<int>& nums, int l, int h, int t) {
int m = 0;
while (l < h) {
m = l + (h - l) / 2;
if (nums[m] < t) {
l = m + 1;
} else {
h = m;
}
}
return l;
}
int search(vector<int>& nums, int target) {
if (nums.size() == 0) return -1;
if (nums.size() == 1) return nums[0] == target ? 0 : -1;
int p = findP(nums);
if (p == -1) {
p = findT(nums, 0 , nums.size() - 1, target);
} else if (nums[0] <= target) {
p = findT(nums, 0, p, target);
} else {
p = findT(nums, p + 1, nums.size() - 1, target);
}
return nums[p] == target ? p : -1;
}
};
复杂度分析
时间复杂度:O(\log n)
空间复杂度:O(1)
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