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Description
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.
Notice
You may assume no duplicate exists in the array.
Example
Given [4, 5, 6, 7, 0, 1, 2]
return 0
Solution
这道题也是经典二分法例题的变形,需要找到第一个下降的元素。图形化数据,相当于一个上升的直线,然后突然掉下来, 再次上升,但是比第一个元素小。target应取最后一个元素的值而不是第一个,因为想把它转换成第一个满足(或不满足)某一条件的问题,及OOXX的问题。如果取target为第一个元素,条件为 First Position <= target 的话,就变成了OXX...XOO...OO,若条件是 First Position < target,则单调增的数组就没有比第一个元素小的数了,得不到任何结果。因此 target 应为最后一个元素,条件是 First Position <= target。
Java
public class Solution {
/**
* @param nums: a rotated sorted array
* @return: the minimum number in the array
*/
public int findMin(int[] nums) {
// write your code here
if (nums == null || nums.length == 0) {
return -1;
}
int start = 0, end = nums.length - 1;
//xxxx
// oooo
int target = nums[nums.length - 1];
// find the first element <= target
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];
}
}
}