LeetCode033- Search in Rotated S
2018-12-26 本文已影响0人
Kay_Coding
Search in Rotated Sorted Array
Question:
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]).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
Your algorithm's runtime complexity must be in the order of O(log n).
Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
解法代码
public class LeetCode33 {
/**
* 二分查找,递归
* 理清思路,列举所有可能的情况,
* 每一次二分可能有一半是有序集合一半是反转集合,也有可能两边都是有序集合
* @param nums
* @param target
* @return
*/
public int search(int[] nums, int target) {
// 空数组,直接返回-1
if(nums == null || nums.length == 0) {
return -1;
}
return search(nums, 0, nums.length - 1, target);
}
private int search(int[] nums, int start, int end ,int target) {
if(nums[start] == target) {
return start;
}
if(nums[end] == target) {
return end;
}
// 未找到结果
if(end - start <= 1) {
return -1;
}
int mid = (start + end + 1) / 2;
if(nums[mid] == target) {
return mid;
}
// 结果在左半部分有序集中
if(nums[start] < nums[mid] && target > nums[start] && target < nums[mid]) {
return search(nums, start, mid, target);
}
// 结果在右半部分有序集中
if(nums[mid] < nums[end] && target > nums[mid] && target < nums[end]) {
return search(nums, mid, end, target);
}
// 结果在左半部分反转集中
if(nums[start] > nums[mid]) {
return search(nums, start, mid, target);
}
// 结果在有半部分反转集中
return search(nums, mid, end, target);
}
}
解法代码
import org.junit.Before;
import org.junit.Test;
import org.junit.runner.RunWith;
import org.junit.runners.Parameterized;
import org.junit.runners.Parameterized.Parameters;
import static org.junit.Assert.assertEquals;
import java.util.Arrays;
import java.util.Collection;
import com.kay.pro.alog.LeetCode33;
@RunWith(Parameterized.class)
public class LeetCode33Test {
private LeetCode33 leetCode;
private int[] nums;
private int target;
private int index;
public LeetCode33Test(int[] nums, int target, int index) {
this.nums = nums;
this.target = target;
this.index = index;
}
@Before
public void before() {
leetCode = new LeetCode33();
}
@Parameters
public static Collection<?> reverserArrays() {
return Arrays.asList(new Object[][]{
{new int[]{4, 5, 6, 7, 0, 1, 2}, 0, 4},
{new int[]{4, 5, 6, 7, 0, 1, 2}, 3, -1},
{new int[]{4, 5, 6, 7, 0, 1, 2}, 6, 2},
{new int[]{8, 7, 6, 5, 4, 3, 2, 1, 0}, 6, 2}
});
}
@Test
public void leetCode33() {
assertEquals(index, leetCode.search(nums, target));
}
}
Output:
Time And Space Complexity
Time: 二分查找时间复杂度
Space: 不需要使用额外的存储空间,原地替换
Tips
二分查找,递归
理清思路,列举所有可能的情况,每一次二分可能有一半是有序集合一半是反转集合,也有可能两边都是有序集合