LeetCode 33-Search in Rotated Sorted Array
LeetCode 33-Search in Rotated Sorted Array
Dylan Liu
发布于 2019-07-01 11:47:42
发布于 2019-07-01 11:47:42
Problem description
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: -1Solution Self
看到这个题目，头脑中第一个蹦出来的就是找到rotated的元素，然后两边分别做binary search就可以了。
第一版代码
class Solution {
public int search(int[] nums, int target) {
if (nums.length == 0) {
return -1;
}
if (nums.length == 1){
return nums[0] == target ? 0 : -1;
}
int pivot = binarySearchPivot(nums, 0, nums.length);
if (pivot == -1){
return searchIndex(nums, 0, nums.length, target);
}
int index = searchIndex(nums, 0, pivot, target);
if (index != -1) {
return index;
}
return searchIndex(nums, pivot, nums.length, target);
}
private int searchIndex(int[] nums, int low, int high, int target) {
if (high <= low) {
return -1;
}
int middle = (low + high) /2;
if (nums[middle] == target) {
return middle;
}
if (nums[middle] < target) {
return searchIndex(nums, middle + 1, high, target);
}
return searchIndex(nums, low, middle, target);
}
private int binarySearchPivot(int[] nums, int low, int high) {
if (high <= low) {
return -1;
}
int middle = (low + high) / 2;
if (middle > 0
&& nums[middle] < nums[middle - 1]) {
return middle;
}
int leftSearch = binarySearchPivot(nums, low, middle);
if (leftSearch != -1) {
return leftSearch;
}
return binarySearchPivot(nums, middle + 1, high);
}
}分两步： 1：找到pivot，就是rotated的那个点， 2：左右两侧分别使用binarySearch来寻找元素，找到返回index，未找到返回-1
提交到LeetCode，Accepted！
不过上面这个程序有一个问题，问题描述中要求算法的复杂度需为log(n)，在搜索pivot的过程中，一直是先搜索左侧，再搜索右侧，这导致当pivot在最右侧的时候，这个binarySearchpivot算法的复杂度为O(n),很明显LeetCode没有强大到计算你程序的复杂度。
改进一版：
package com.dylan.leetcode;
import org.junit.Assert;
import org.junit.Test;
/**
* Created by liufengquan on 2018/6/20.
*/
public class SearchInRotatedSortedArray {
public int search(int[] nums, int target) {
if (nums.length == 0) {
return -1;
}
if (nums.length == 1){
return nums[0] == target ? 0 : -1;
}
int pivot = binarySearchPivot(nums, 0, nums.length);
if (pivot == -1){
return searchIndex(nums, 0, nums.length, target);
}
int index = searchIndex(nums, 0, pivot, target);
if (index != -1) {
return index;
}
return searchIndex(nums, pivot, nums.length, target);
}
private int searchIndex(int[] nums, int low, int high, int target) {
if (high <= low) {
return -1;
}
int middle = (low + high) /2;
if (nums[middle] == target) {
return middle;
}
if (nums[middle] < target) {
return searchIndex(nums, middle + 1, high, target);
}
return searchIndex(nums, low, middle, target);
}
private int binarySearchPivot(int[] nums, int low, int high) {
if (high <= low) {
return -1;
}
int middle = (low + high) / 2;
if (middle > 0
&& nums[middle] < nums[middle - 1]) {
return middle;
}
if (nums[middle] <= nums[high - 1]) {
return binarySearchPivot(nums, low, middle);
}
return binarySearchPivot(nums, middle + 1, high);
}
@Test
public void test() {
SearchInRotatedSortedArray searchInRotatedSortedArray = new SearchInRotatedSortedArray();
int[] nums = new int[]{4, 5, 6, 7, 0, 1, 2};
Assert.assertTrue( searchInRotatedSortedArray.search(nums, 0) == 4);
Assert.assertTrue(searchInRotatedSortedArray.search(nums, 3) == -1);
nums = new int[]{5, 4};
org.junit.Assert.assertTrue(searchInRotatedSortedArray.binarySearchPivot(nums, 0, 2) == 1);
org.junit.Assert.assertTrue(searchInRotatedSortedArray.search(nums, 4) == 1);
org.junit.Assert.assertTrue(searchInRotatedSortedArray.search(nums, 5) == 0);
nums = new int[]{1, 3};
Assert.assertTrue(searchInRotatedSortedArray.binarySearchPivot(nums, 0, 2) == -1);
Assert.assertTrue(searchInRotatedSortedArray.search(nums, 0) == -1);
nums = new int[]{3, 4, 5, 6, 1, 2};
System.out.println(searchInRotatedSortedArray.binarySearchPivot(nums, 0, nums.length));
Assert.assertTrue(searchInRotatedSortedArray.binarySearchPivot(nums, 0, nums.length) == 4);
Assert.assertTrue(searchInRotatedSortedArray.search(nums, 2) == nums.length -1);
}
}基于pivot元素的左右两侧都比它大的原则，以及pivot左右都是升序的题设，检查了一下，对于检测middle左侧还是右侧有一个判定，这样程序的执行时间复杂度就是O(log(n))了。
Solution Others
其实在寻找pivot的过程中我们已经看到了，对于搜索左侧还是右侧我们是可以根据middle跟high的元素大小来判定出来的，因此这个pivot其实没有必要搜索出来，直接根据target的值做二分搜索就可以了。看了一下Discuss里面的网友上传解决方法，一般都是用这种。
在这贴一个他人的解决方法：
class Solution {
public int search(int[] nums, int target) {
int start = 0;
int end = nums.length - 1;
while (start <= end) {
int mid = (start + end) / 2;
if (nums[mid] == target) return mid;
if (nums[mid] > target) {
if (nums[start] <= nums[mid] && nums[end] <= nums[start] && nums[start] > target)
start = mid + 1;
else end = mid - 1;
}
else {
if (nums[mid] <= nums[end] && nums[end] <= nums[start] && nums[end] < target)
end = mid - 1;
else start = mid + 1;
}
}
return -1;
}
}DONE
考察要点：二分搜索
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