【LeetCode】Find Minimum in Rotated Sorted Array 在旋轉數組中找最小數

Add Date 2014-10-15

Find Minimum in Rotated Sorted Array

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.

這個很簡單，由於沒有重複數字，數組本質上仍是有序的，用相似二分查找的方法複雜度O(log n)。記得考慮一下整個數組沒有 rotated 的狀況。

20號又加了有重複數字的題II，直接附II的 code 吧，II中須要考慮前中後三個數相等的狀況，此時沒法肯定最小值在哪邊，只能遍歷一遍。

 1 class Solution {
 2 public:
 3     int findMinN(vector<int> &num, int min, int max) {
 4         int minN = num[min];
 5         for(int i = min+1; i <= max; ++i) {
 6             if(num[i] < minN)
 7                 minN = num[i];
 8         }
 9         return minN;
10     }
11     int findMin(vector<int> &num, int min, int max) {
12         if(min == max || num[min] < num[max])
13             return num[min];
14         int mid = (max+min)>>1;
15         if(num[mid] < num[min]) {
16             return findMin(num, min, mid);
17         }
18         else if(num[mid] > num[max]) {
19             return findMin(num, mid+1, max);
20         }
21         else
22             return findMinN(num, min, max);
23     }
24     int findMin(vector<int> &num) {
25         int len = num.size();
26         return findMin(num, 0, len-1);
27     }
28 };