[LeetCode] 154. Find Minimum in Rotated Sorted Array II

轉自LeetCode

Suppose an array sorted in ascending order 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.

<Solution>
這題是 Find Minimum in Rotated Sorted Array 的衍生題

不難，只改動一個地方

因為現在會有重複的值出現，所以判斷是從 if(nums.front() > nums.back())

變成 if(nums.front() >= nums.back())

code 如下
c++

	class Solution {
	public:
	int findMin(vector<int>& nums) {
	if(nums.front() >= nums.back()) {
	//>> rotated
	for(int i = nums.size() - 1; i >= 1; i--) {
	if(nums[i-1] > nums[i]) {
	return nums[i];
	}
	}
	}
	//>> not rotated
	return nums[0];
	}
	};

view raw findMinInRotatedArrayII.cpp hosted with ❤ by GitHub

kotlin

	class Solution {
	fun findMin(nums: IntArray): Int {
	if (nums.first() >= nums.last()) {
	for (i in nums.lastIndex downTo 1) {
	if (nums[i-1] > nums[i]) {
	return nums[i]
	}
	}
	}
	return nums[0]
	}
	}

view raw find_minimum_in_rotated_sorted_array_II.kt hosted with ❤ by GitHub