标签：style blog ar color os sp for div log
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.
对数组A,
1)如果A[mid] < A[low], 说明minimum在[mid, low]之间。(存在rotated)
2)若A[mid] > A[low], 有两种情况：
a)A[mid] > A[high], minimum在[mid, high]之间。(存在rotated)
b)A[mid] < A[high], minimum在[low, mid]之间。(不存在rotated)
注意终止条件：A[mid]小于左右两侧或者low == mid
1 int findMin(vector<int> &num) 2 { 3 int low = 0, high = num.size() - 1, mid = 0; 4 while (low <= high) 5 { 6 mid = (low + high) / 2; 7 if ((mid == 0 || num[mid] < num[mid - 1]) && (mid == num.size() - 1 || num[mid] < num[mid + 1])) 8 break; 9 if (num[mid] < num[low]) 10 { 11 high = mid - 1; 12 } 13 else if (num[mid] > num[low]) 14 { 15 if (num[mid] > num[high]) 16 low = mid + 1; 17 else if (num[mid] < num[high]) 18 high = mid - 1; 19 } 20 else 21 { 22 return min(num[low], num[high]); 23 } 24 } 25 26 return num[mid]; 27 }
leetcode. Find Minimum in Rotated Sorted Array
标签：style blog ar color os sp for div log
原文地址：http://www.cnblogs.com/ym65536/p/4150061.html
