Leetcode 153. 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.

解题思路：

方法一：推荐用binary search. O(logn)

(1) A[mid] < A[end]：A[mid : end] sorted => min不在A[mid+1 : end]中
搜索A[start : mid]
(2) A[mid] > A[end]：A[start : mid] sorted且又因为该情况下A[end]<A[start] => min不在A[start : mid]中
搜索A[mid+1 : end]
(3) base case：
a. start = end，必然A[start]为min，为搜寻结束条件。
b. start + 1 = end，此时A[mid] = A[start]，而min = min(A[mid], A[end])。而这个条件可以合并到(1)和(2)中。
画图解决，直接明了。

方法二：直接观察发现最小值就是某值比前面的那个数小，就是最小值。也对。当然复杂度是O(n). 还是方法一更好。 直接不用方法二，本题考的就是binary search.

Java code:

1. Binary search

public class Solution {
public int findMin(int[] nums) {
int left = 0, right = nums.length-1;
while(left < right) {
int mid = left + (right - left) / 2;
if(nums[mid] < nums[right]){
right = mid;
}else{
left = mid+1;
}
}
return nums[left];
}
}

1.2 九章算法重做本题 2016.01.18

public class Solution {
public int findMin(int[] nums) {
if(nums == null || nums.length == 0) {
return -1;
}
int start = 0, end = nums.length-1;
int target = nums[end];
while(start + 1 < end) {
int mid = start + (end - start) / 2;
if(nums[mid] <= target) {
end = mid;
}else {
start = mid;
}
}
if(nums[start] <= target) {
return nums[start];
}else {
return nums[end];
}
}
}

2.

public class Solution {
public int findMin(int[] nums) {
for(int i = 1; i < nums.length; i++){
if(nums[i] < nums[i-1]){
return nums[i];
}
}
return nums[0];
}
}

Reference:

1. http://bangbingsyb.blogspot.com/2014/11/leecode-find-minimum-in-rotated-sorted.html