[leetcode]Find Minimum in Rotated Sorted Array

最新推荐文章于 2024-11-13 08:49:23 发布

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本文介绍了解决旋转排序数组中最小元素问题的方法，包括使用二分查找算法进行优化，确保在不重复的情况下找到最小值。

From : https://leetcode.com/problems/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.

class Solution {
public:
    int findMin(vector<int>& nums) {
        if(nums.front()<nums.back()) return nums[0];
        int start=0, end=nums.size()-1;
        while(start<end) {
            if(start+1==end) return min(nums[start], nums[end]);
            int mid = (start+end)>>1;
            if(nums[mid]==nums[start] && nums[mid]==nums[end]) return check(nums, start, end);
            if(nums[mid]>=nums[start]) {
                start = mid;
            } else {
                end = mid;
            }
        }
        return nums[start];
    }
    int check(vector<int>& nums, int start, int end) {
        int min = nums[start];
        for(int i=start+1; i<=end;i++) {
            if(nums[i]<min) min=nums[i];
        }
        return min;
    }
};

public class Solution {
    public int findMin(int[] nums) {
        if (nums == null) {
			return 0;
		}
		int l = 0, r = nums.length - 1;
		int ans = nums[0];
		while (l <= r) {
			if (nums[l] < nums[r]) {
				return ans < nums[l] ? ans : nums[l];
			}
			int m = (l + r) >> 1;
			if (nums[m] < ans) {
				ans = nums[m];
			}
			if (nums[m] >= nums[l]) {
				l = m + 1;
			} else {
				r = m - 1;
			}
		}
		return ans;
    }
}