两道题

33. Search in Rotated Sorted Array

https://leetcode.com/problems/search-in-rotated-sorted-array/

81. Search in Rotated Sorted Array II

https://leetcode.com/problems/search-in-rotated-sorted-array-ii/

这两道题都是属于循环有序数组的查找问题，无论是查找具体元素的下标还是检查一个目标值是否在数组中，主要的判断逻辑都是一致的。

思路

这种将有序数组翻转成循环有序数组的解决办法是将原数组分段。用首元素start、中间元素mid、尾元素end，可以将数组分为两个子数组s1，s2。

则这两个子数组中，必然有至少一个子数组是有序的。

那么如何确定那一段是有序的呢，通过分析可以看到只有3种情况：

当start就是A中最小的元素时，以下不等式成立：start <= mid <= end
当最小值位于(start, mid]时，则有：mid <= end <= start
当最小值出现在(mid，end]时，则有：end <= start <= mid

所以通过start, mid, end的大小关系，就可以判断出s1和s2哪个是有序的。

通过比较要查找的目标target与start，mid，end的大小关系，就可以知道target位于哪个子数组。

若target位于有序的子数组，则用二分查找就可以了。否则，对无序的子数组重复刚才的过程就可以了。

代码

package com.lingyejun.dating.chap11.toutiao;public class RotatedArrayQuery {/*** 33. Search in Rotated Sorted Array* 查找目标值下标** @param nums* @param target* @return*/public int searchTargetIndex(int[] nums, int target) {int start = 0;int end = nums.length - 1;while (start <= end) {// (start ... middle ... end)int mid = (start + end) / 2;if (nums[mid] == target) {return mid;}// 前半段数组是有序的if (nums[start] < nums[mid]) {// target在前半段数组中if (target < nums[mid] && target >= nums[start]) {// 将原数组折半，取出前半段数组end = mid - 1;} else {// target在后半段数组中start = mid + 1;}}// 后半段数组是有序的else if (nums[mid] < nums[end]) {// target在后半段数组中if (target > nums[mid] && target <= nums[end]) {// 将原数组折半，取出后半段数组start = mid + 1;} else {// target在前半段数组中end = mid - 1;}} else {// 此种场景{1, 0, 1, 1, 1} start = mid = endstart++;}}return -1;}/*** 81. Search in Rotated Sorted Array II** 查找是否存在目标值** @param nums* @param target* @return*/public static boolean searchExistsKey(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 true;}if (nums[start] < nums[mid]) {if (nums[start] <= target && target < nums[mid]) {end = mid - 1;} else {start = mid + 1;}} else if (nums[start] > nums[mid]) {if (nums[mid] < target && target <= nums[end]) {start = mid + 1;} else {end = mid - 1;}} else {start++;}}return false;}public static void main(String[] args) {int[] array = new int[]{1, 0, 1, 1, 1};RotatedArrayQuery caq = new RotatedArrayQuery();System.out.println(caq.searchTargetIndex(array, 0));System.out.println(caq.searchExistsKey(array, 1));}}

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Search in Rotated Sorted Array - 循环有序数组查找问题

两道题

33. Search in Rotated Sorted Array

81. Search in Rotated Sorted Array II

思路

代码

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