LeetCode 题解（62）: Find Minimum in Rotated Sorted Array

来源：互联网发布：js 对象转换为url参数编辑：程序博客网时间：2024/06/07 02:19

题目：

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.

题解：最小值永远出现在未排好序的一半。若low < mid < high，则两半都未排好序的状态，最小值为low。若只剩两个元素，最小值为两者中的小的。

C++版：

class Solution {public:    int findMin(vector<int> &num) {        int low = 0, high = num.size()-1;        if(low == high)            return num[low];        while(low <= high) {            int mid = (low + high) / 2;            if(num[mid] > num[low] && num[mid] > num[high])                 low = mid;            else if(num[mid] < num[low] && num[mid] < num[high]) {                high = mid;            } else if(num[mid] > num[low] && num[mid] < num[high]) {                return num[low];            } else if(num[mid] == num[low]) {                if(num[low] < num[high])                    return num[low];                else                    return num[high];            }        }    }};

Java版：

public class Solution {    public int findMin(int[] num) {        int low = 0, high = num.length-1;                if(num.length == 1)            return num[0];                while(low <= high) {            int mid = (low + high) / 2;            if(num[mid] > num[low] && num[mid] < num[high])                 return num[low];            else if(num[mid] > num[low] && num[mid] > num[high])                low = mid;            else if(num[mid] < num[low] && num[mid] < num[high])                high = mid;            else {                if(num[low] < num[high])                    return num[low];                else                    return num[high];            }        }        return 0;    }}

Python版：

class Solution:    # @param num, a list of integer    # @return an integer    def findMin(self, num):        if len(num) == 1:            return num[0]                    low = 0        high = len(num)-1        while low <= high:            mid = (low + high) / 2            if num[low] < num[mid] and num[mid] < num[high]:                return num[low]            elif num[low] < num[mid] and num[high] < num[mid]:                low = mid            elif num[low] > num[mid] and num[high] > num[mid]:                high = mid            else:                if num[low] < num[high]:                    return num[low]                else:                    return num[high]

0 0