LeetCode153. Find Minimum in Rotated Sorted Array

文章目录

• [153. Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)
• [154. Find Minimum in Rotated Sorted Array II](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/)

153. Find Minimum in Rotated Sorted Array

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.

You may assume no duplicate exists in the array.

Example 1:

Input: `[3,4,5,1,2]`
Output: 1

Example 2:

Input: `[4,5,6,7,0,1,2]`
Output: 0

---

题目：旋转（有序）数组中的最小值，假设数组中无重复元素。

思路：二分查找，参见Compact and clean C++ solution。对于旋转数组[4,5,6,7,0,1,2]而言，最小值落在后半段有序数组的首位。nums[start] > nums[end]；对于mid = start + (end - start)/2，若是nums[start] > nums[mid]，说明mid落在了后半段有序的数组中，那么最小值落在[start, mid]之间，若是nums[start] < nums[mid]，说明mid落在了前半段有序的数组中，那么最小值在[mid, end]之中。

工程代码下载

class Solution {
public:
    int findMin(vector<int>& nums) {
        int r = nums.size() - 1;
        int l = 0;
        while(l < r){
            if(nums[l] < nums[r]) //数组已经有序
                return nums[l];
            int mid = l + (r - l)/2;
            if(nums[l] > nums[mid])
                r = mid;
            else
                l = mid + 1;
        }
        return nums[l];
    }
};

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

---

154. Find Minimum in Rotated Sorted Array II

若是数组中存在重复元素。

https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/discuss/48808/My-pretty-simple-code-to-solve-it

class Solution {
public:
    int findMin(vector<int> &num) {
        int lo = 0;
        int hi = num.size() - 1;
        int mid = 0;

        while(lo < hi) {
            mid = lo + (hi - lo) / 2;

            if (num[mid] > num[hi]) {
                lo = mid + 1;
            }
            else if (num[mid] < num[hi]) {
                hi = mid;
            }
            else { // when num[mid] and num[hi] are same
                hi--;
            }
        }
        return num[lo];
    }
};

When num[mid] == num[hi], we couldn’t sure the position of minimum in mid's left or right, so just let upper bound reduce one.