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Be carefully with one element and two element sitution.
1. Since mid = (start + end) / 2 is alway shifting to the left. So when we do comparision, not use num[mid] > num[start] but num[mid] > num[end].
2. We need find the minimum, So when we are closing to the minimum, we need get start = mid + 1 and end = mid as offset for left shifting.
1 class Solution { 2 public: 3 int findMin(vector<int> &num) { 4 int start = 0, end = num.size()-1, mid = 0; 5 if (num[start] < num[end]) return num[start]; 6 while (start < end) { 7 mid = (start + end)/2; 8 if (num[mid] > num[end]) { 9 start = mid + 1; 10 } else { 11 end = mid; 12 } 13 } 14 return num[start]; 15 } 16 };
LeetCode – Refresh – Find Minimum in Rotated Array
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原文地址：http://www.cnblogs.com/shuashuashua/p/4349503.html
