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Copy path33.search-in-rotated-sorted-array.cpp
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93 lines (81 loc) · 2.09 KB
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/**
* 33. Search in Rotated Sorted Array
*
* 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
*/
#include "testharness.h"
class Solution {
public:
int search(vector<int>& nums, int target) {
return binarySearch(nums, 0, nums.size() - 1, target);
}
int binarySearch(const vector<int>& N, int l, int h, int t) {
// std::cout << l << "\t" << h << "\n";
if (h - l < 5) {
for (int i = l; i <= h; i++) {
if (N[i] == t)
return i;
}
return -1;
}
int m = (h+l)/2;
if (N[m] == t)
return m;
if (N[l] < N[h]) {
if (t < N[l] || t > N[h])
return -1;
else if (t < N[m])
return binarySearch(N, l, m-1, t);
else
return binarySearch(N, m+1, h, t);
}
// here we have N[l] > N[h]
if (t > N[h] && t < N[l])
return -1;
if (N[l] < N[m]) {
if (t >= N[l] && t < N[m])
return binarySearch(N, l, m-1, t);
else
return binarySearch(N, m+1, h, t);
}
if (t <= N[h] && t > N[m])
return binarySearch(N, m+1, h, t);
else
return binarySearch(N, l, m-1, t);
}
};
TEST(Solution, test0) {
vector<int> nums = {6,7,1,2,3,4,5};
ASSERT_EQ(0, search(nums, 6));
}
TEST(Solution, test) {
vector<int> nums = {4,5,6,7,0,1,2};
ASSERT_EQ(6, search(nums, 2));
ASSERT_EQ(4, search(nums, 0));
ASSERT_EQ(-1, search(nums, 3));
}
TEST(Solution, test2) {
vector<int> nums = {0,1,2,4,5,6,7};
ASSERT_EQ(3, search(nums, 4));
ASSERT_EQ(-1, search(nums, 3));
}
TEST(Solution, test3) {
vector<int> nums = {3,1};
ASSERT_EQ(1, search(nums, 1));
}