L159 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 become4 5 6 7 0 1 2
).
Find the minimum element.
Notice
You may assume no duplicate exists in the array.
Example
Given[4, 5, 6, 7, 0, 1, 2]
return0
还是2分模板,这个没重复。还是找有序的一半,然后判断下标移动。想像一条斜线砍一半。竖线表示mid的位置,如果砍在左边,min在右边,如果砍在右边,min在左边。

public int findMin(int[] nums) {
if (nums == null || nums.length == 0) {
return -1;
}
int start = 0;
int end = nums.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
//前半有序,找后半,因为中间本应该小于后面的,所以min在后半。
if (nums[mid] > nums[end]) {
start = mid;
} else {//后半有序找前半
end = mid;
}
}
if (nums[start] < nums[end]) {
return nums[start];
} else {
return nums[end];
}
}
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