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[LeetCode] Find Minimum in Rotated Sorted Array II
Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?Would this affect the run-time complexity? How and why?
Suppose a sorted array 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).
Find the minimum element.
The array may contain duplicates.
跟上一题一样，注意相等的情况要特殊考虑。
1 class Solution { 2 public: 3 int findMinHelper(vector<int> &num, int left, int right) { 4 int mid = (left + right) / 2; 5 if (mid == left || mid == right) { 6 return min(num[left], num[right]); 7 } 8 if (num[mid] < num[right]) { 9 return findMinHelper(num, left, mid);10 } else if (num[mid] > num[right]) {11 return findMinHelper(num, mid, right);12 } else {13 return findMinHelper(num, left, right - 1);14 }15 }16 int findMin(vector<int> &num) {17 return findMinHelper(num, 0, num.size() - 1);18 }19 };
或者：
1 class Solution { 2 public: 3 int findMinHelper(vector<int> &num, int left, int right) { 4 int mid = (left + right) / 2; 5 if (mid == left || mid == right) { 6 return min(num[left], num[right]); 7 } 8 if (num[mid] < num[right]) { 9 return findMinHelper(num, left, mid);10 } else if (num[mid] > num[right]) {11 return findMinHelper(num, mid, right);12 } else {13 int res = num[left];14 for (int i = left; i <= right; ++i) {15 res = min(res, num[i]);16 }17 return res;18 }19 }20 int findMin(vector<int> &num) {21 return findMinHelper(num, 0, num.size() - 1);22 }23 };
[LeetCode] Find Minimum in Rotated Sorted Array II
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