LeetCode Find Minimum in Rotated Sorted Array II 有条件的二分
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思路:
有条件的二分。
若相等,则走一步,此时不二分。
class Solution {public: int findMin(vector<int>& nums) { int n = nums.size(); int left = 0; int right = n - 1; while(left <= right) { int mid = left + (right - left) / 2; if(nums[mid] > nums[right]) { left = mid + 1; }else if(nums[mid] == nums[right]) { right--; }else { right = mid; } } return nums[left]; }};
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