【LeetCode】Find Minimum in Rotated Sorted Array 解题报告

【题目】

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.

【解析】

The input could either be rotated [4, 5, 6, 7, 0, 1, 2] or not rotated [0, 1, 2, 4, 5, 6, 7] or totally rotated. Binary search can be used to solve this question because input is sorted in some way. But, the point is, when cutting half and then determining
the answer range, do we compare nums[middle] with nums[start] or nums[end] or both are OK? Let's discuss which one of them can be applied to all input cases.

1. Compared nums[middle] with nums[start]

when input is rotated: eg. [4, 5, 6, 7, 0, 1, 2]. 

if nums[middle] > nums[start], then index of minimum is between [middle, end]; else [start, middle].

When input is not rotated: eg. [0, 1, 2, 4, 5, 6, 7]. nums[middle] > nums[start], however index of minimum is not between [middle, end]

When input is totally rotated: eg [7, 6, 5, 4, 2, 1, 0]. nums[middle] < nums[start], however index of minimum is not between [start, middle]

So if compared nums[middle] with nums[start], code can not apply to second and third cases.

2. Compared nums[middle] with nums[end]

when input is rotated: eg. [4, 5, 6, 7, 0, 1, 2]. 

if nums[middle] > nums[end], then index of minimum is between [middle, end]; else [start, middle].(same way as way 1 case 1)

When input is not rotated: eg. [0, 1, 2, 4, 5, 6, 7]. nums[middle] < nums[end], [start, middle]? BINGO

When input is totally rotated: eg [7, 6, 5, 4, 2, 1, 0]. nums[middle] > nums[end], [middle, end]? BINGO

So code can apply to all three cases.

【代码】

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