标签：get fir sub dup als sts exit hand log
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
this is my first code, it can‘t satisfy the last condition:
class Solution {
public:
int search(vector<int>& nums, int target) {
int len = nums.size();
if (len == 0) return -1;
bool flag = false;
if (target < nums[0]) {
for (int i = len-1; i > 0; --i) {
if (nums[i] == target) {
flag = true;
return i;
}
}
} else {
for (int i = 0; i < len; ++i) {
if (nums[i] == target) {
flag = true;
return i;
}
}
}
if (!flag) {
return -1;
}
}
};
this is the right way to use binary search:
class Solution {
public:
int search(vector<int>& nums, int target) {
int len = nums.size();
if (len == 0) return -1;
int l = 0;
int r = len - 1;
while (l <= r) {
int m = (l + r) / 2;
if (nums[m] == target) return m;
if (nums[m] > nums[r]) {
if (target < nums[m] && target >= nums[l]) {
r = m - 1;
} else {
l = m + 1;
}
} else if (nums[m] < nums[l]) {
if (target <= nums[r] && target > nums[m]) {
l = m + 1;
} else {
r = m -1;
}
} else {
if (target < nums[m]) {
r = m -1;
} else {
l = m + 1;
}
}
}
return -1;
}
};
Runtime: 8 ms, faster than 25.99% of C++ online submissions for Search in Rotated Sorted Array.
but this way cost more time before the previous way.
33. Search in Rotated Sorted Array
标签：get fir sub dup als sts exit hand log
原文地址：https://www.cnblogs.com/ruruozhenhao/p/9775497.html
