Find Minimum in Rotated Sorted Array II

Question

Problem Statement

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.

The array may contain duplicates.

Example

Given [4,4,5,6,7,0,1,2] return 0

题解

由于此题输入可能有重复元素,因此在num[mid] == num[end]时无法使用二分的方法缩小start或者end的取值范围。此时只能使用递增start/递减end逐步缩小范围。

C++

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class Solution {
public:
/**
* @param num: a rotated sorted array
* @return: the minimum number in the array
*/
int findMin(vector<int> &num) {
if (num.empty()) {
return -1;
}

vector<int>::size_type start = 0;
vector<int>::size_type end = num.size() - 1;
vector<int>::size_type mid;
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (num[mid] > num[end]) {
start = mid;
} else if (num[mid] < num[end]) {
end = mid;
} else {
--end;
}
}

if (num[start] < num[end]) {
return num[start];
} else {
return num[end];
}
}
};

Java

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public class Solution {
/**
* @param num: a rotated sorted array
* @return: the minimum number in the array
*/
public int findMin(int[] num) {
if (num == null || num.length == 0) return Integer.MIN_VALUE;

int lb = 0, ub = num.length - 1;
// case1: num[0] < num[num.length - 1]
// if (num[lb] < num[ub]) return num[lb];

// case2: num[0] > num[num.length - 1] or num[0] < num[num.length - 1]
while (lb + 1 < ub) {
int mid = lb + (ub - lb) / 2;
if (num[mid] < num[ub]) {
ub = mid;
} else if (num[mid] > num[ub]){
lb = mid;
} else {
ub--;
}
}

return Math.min(num[lb], num[ub]);
}
}

源码分析

注意num[mid] > num[ub]时应递减 ub 或者递增 lb.

复杂度分析

最坏情况下 $$O(n)$$, 平均情况下 $$O(\log n)$$.


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