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_153_Find_Minimum_in_Rotated_Sorted_Array.java
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_153_Find_Minimum_in_Rotated_Sorted_Array.java
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package com.github.junyu.solution.leetCode.medium.array;
public class _153_Find_Minimum_in_Rotated_Sorted_Array {
/* 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.
Example 1:
Input: [3,4,5,1,2]
Output: 1
Example 2:
Input: [4,5,6,7,0,1,2]
Output: 0*/
/**
* 找出数组中的最小值,原先的数组是完全升序,但是被旋转一次之后,就可能变成了两个升序数组
* 估计这个性质,那么旋转后的数组中第1个元素必定大于最后一个元素,这题可以使用二分查找去实现
* 题目限定了没有重复元素,那么当出现第一个元素小于最后一个元素时,就说明此时数组已经是一个排过序的数组,直接返回midIndex
* 在二分查找的过程中,只要左边的指针所指的值大于右边,我们就去查找。只有当范围缩小到right-left=1的时候,就说明left已经为序第一个升序
* 数组的最后一个元素,而right指针在第二个升序子数组的第一个元素。那么right所指即使最小的元素。
* 缩小范围的时候,无论的左还是右,仍然要继续考察mid位置的元素。
* @param nums
* @return
*/
public int findMin(int[] nums) {
if (nums == null || nums.length == 0)
return 0;
int left = 0;
int right = nums.length - 1;
int midIndex = left;
while (nums[left] > nums[right]) {
if (right - left == 1) {
midIndex = right;
break;
}
int mid = (right - left) / 2 + left;
if (nums[mid] > nums[left])
left = mid;
else if (nums[mid] < nums[right])
right = mid;
}
return nums[midIndex];
}
public static void main(String[] args) {
System.out.println(new _153_Find_Minimum_in_Rotated_Sorted_Array().findMin(new int[]{3, 4, 5, 1, 2}));
System.out.println(new _153_Find_Minimum_in_Rotated_Sorted_Array().findMin(new int[]{4, 5, 6, 7, 0, 1, 2}));
}
}