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.
You may assume no duplicate exists in the array.
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) return min(num[left], num[right]); 6 if (num[mid] < num[right]) { 7 return findMinHelper(num, left, mid); 8 } else { 9 return findMinHelper(num, mid, right); 10 } 11 } 12 13 int findMin(vector<int> &num) { 14 return findMinHelper(num, 0, num.size() - 1); 15 } 16 };
二分查找,注意边界情况的判断。用中间元素与最左边元素比较。只有两种情况,最小值要么在最左边,要么在最右边。二分时要保留中间元素,以防止最小值就是中间的元素。