Leetcode Find Minimum in Rotated Sorted Array II

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.

The array may contain duplicates.

解题思路：

与Leetcode Find Minimum in Rotated Sorted Array 的区别只是有重复数字。

方法一：推荐用binary search. O(logn)

当A[mid] = A[end]时，无法判断min究竟在左边还是右边。

但可以肯定的是可以排除A[end]：因为即使min = A[end]，由于A[end] = A[mid]，排除A[end]并没有让min丢失。所以增加的条件是：
A[mid] = A[end]：搜索A[start : end-1]

画图解决

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。代码和之前那题没有任何变化。不用这个方法！

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 if(nums[mid] > nums[right]){
left =  mid+1;
}else {
right--;
}
}
return nums[left];
}
}

1.2 九章算法模板方法 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 mid;
while(start + 1 < end) {
mid = start + (end - start) / 2;
if(nums[mid] < nums[end]) {
end = mid;
}else if (nums[mid] > nums[end]) {
start = mid;
}else {
end--;
}
}
if(nums[start] < nums[end]) {
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