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.
思路分析:這題主要考察二分查找,在Rotated Sorted Array中找最小和Search in Rotated Sorted Array類似,只不過此時需要不斷把最左邊的元素A[l]與A[m]比較
1. 如果A[m] < A[l] 那麼應該在左邊找,更新右邊的邊界right;
2.如果A[m] > A[l] 那麼應該在右邊找,更新左邊的邊界left;
3.如果A[m]= A[l] ,l++;
上述規律可以通過舉例子很快發現,這樣可以每次減少一半元素,得到O(logN)的算法。注意這題A[m] = A[l] 也有可能發生,雖然沒有重複元素,但是m和l可能指向同一個元素,比如當這個數組只有兩個元素的時候。如果A[m]= A[l] ,可以執行l++;丟掉一個“冗餘元素”也不會丟掉最小值。對於Rotated Sorted Array中有冗餘值的情況可以參考本題的擴展Find Minimum in Rotated Sorted Array II。
AC Code
public class Solution {
public int findMin(int[] nums) {
int n = nums.length;
int l = 0;
int r = n-1;
int min = nums[0];
while(l < r){
int m = l + (r-l) / 2;
if(nums[m] < nums[l]) {
min = Math.min(nums[m], min);
r = m;
} else if(nums[m] > nums[l]){
min = Math.min(nums[l], min);
l = m;
} else {
l++;
}
}
min = Math.min(nums[l], min);
min = Math.min(nums[r], min);
return min;
}
}