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March 31, 2018 14:44
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在一个旋转过的数组中搜索一个数,返回它的索引
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class Solution1 { | |
public: | |
/** | |
* @param A: an integer rotated sorted array | |
* @param target: an integer to be searched | |
* @return: an integer | |
*/ | |
int search(vector<int> &A, int target) { | |
int size = A.size(); | |
if (size == 0) { | |
return -1; | |
} | |
int low = 0, high = size-1; | |
return binarySearch(A, target, low, high); | |
} | |
int binarySearch(vector<int> &A, int target, int low, int high) { | |
int mid = (low+high)/2; | |
if (A[mid] == target) { | |
return mid; | |
} | |
if (high < low) { | |
return -1; | |
} | |
// 说明左半边是正常排序的 | |
if (A[low] < A[mid]) { | |
if (target >= A[low] && target <= A[mid]) { | |
// 要找的数字就在这半边 | |
return binarySearch(A, target, low, mid-1); | |
} else { | |
return binarySearch(A, target, mid+1, high); | |
} | |
} else if (A[low] > A[mid]) { | |
// 说明右半边是正常排序的 | |
if (target >= A[mid] && target <= A[high]) { | |
return binarySearch(A, target, mid+1, high); | |
} else { | |
return binarySearch(A, target, low, mid-1); | |
} | |
} else { | |
if (A[mid] != A[high]) { | |
return binarySearch(A, target, mid+1, high); | |
} else { | |
int result = binarySearch(A, target, low, mid-1); | |
if (result == -1) { | |
result = binarySearch(A, target, mid+1, high); | |
return result; | |
} else { | |
return result; | |
} | |
} | |
} | |
return -1; | |
} | |
}; | |
class Solution2 { | |
public: | |
int search(vector<int>& nums, int target) { | |
int size = nums.size(); | |
if (size == 1) { | |
return nums[0] == target?0:-1; | |
} | |
int left = 0,right = size; | |
while(left != right) { | |
int mid = left + (right-left)/2; | |
if (nums[mid] == target) { | |
return mid; | |
} else if(nums[left] <= nums[mid]) { | |
if (nums[left] <= target && target < nums[mid]) { | |
right = mid; | |
} else { | |
left = mid + 1; | |
} | |
} else { | |
if (nums[mid] < target && target <= nums[right-1]) { | |
left = mid + 1; | |
} else { | |
right = mid; | |
} | |
} | |
} | |
return -1; | |
} | |
}; | |
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