https://leetcode-cn.com/problems/search-in-rotated-sorted-array/submissions/
从mid处划分,至少有一半是有序的。如果target在有序部分的区间内,可使用二分查找进行查找,如果没有找到直接返回-1。但如果target不在有序部分的区间内,并不能直接得出没有找到target的结论,因为有可能被旋转到了无序的区间。
每次在有序区间里找target,如果找到返回,如果没有找到,继续把无序区间一分为二,然后在新的有序区间寻找。
package binSearch; public class search_33 { public int search(int[] nums, int target) { int left = 0; int right = nums.length-1; int mid = (left+right)/2; while (left<=right) { mid = (left+right)/2; //左边有序 if (nums[mid]>nums[left]){ //先在有序区间里找 //目标值在有序范围内 if (target<=nums[mid] && target>=nums[left]){ return binSea(nums, left, mid, target); } //有序区间没有再去无序区间找 left = mid+1; }else { //右边有序 if (target>nums[mid]&&target<nums[right]){ return binSea(nums,mid,right,target); } right=mid-1; } } return -1; } public int binSea(int[]nums,int left,int right,int target){ int mid; while (left<=right){ mid = (left+right)/2; if (nums[mid]==target){ return mid; }else if (nums[mid]>target){ right = mid-1; }else { left = mid+1; } } return -1; } public static void main(String[] args) { int[] arr = {4, 5, 6, 7, 0, 1, 2}; int target = 9; search_33 fun = new search_33(); int result = fun.search(arr, target); System.out.println(result); } }
原文:https://www.cnblogs.com/AshOfTime/p/10798237.html