Search in Rotated Sorted Array LeetCode解题报告
2017-08-24 15:49
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题目
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.
思路
可以暴力法找到,这里略过。二分搜索法,对于这种有序的序列查找某个值是个简单同时又高效的方法。这里的数组是部分有序的,使用二分法,总会有一半是有序的,对有序的一半,判断要查找的值是否在这个范围内,若在,就在这个范围内查找,若不在,就去另一个范围内查找。
怎么判断哪一半有序呢?对于左半边,如果中间值大于最左边的值,那么左半边有序,对于右半边,若最优边的值大于中间值,那么右半边有序。
代码
public static int search2(int[] nums, int target) { int len = nums.length; int left = 0,right = len-1; while(left <= right){ int mid = (left+right) / 2; if(target == nums[mid]){ return mid; } //左边是有序的 if(nums[left] <= nums[mid]){ if(target < nums[mid] && target >= nums[left]){ right = mid - 1; }else{//左边有序,但是都大于要搜索值,不在搜索范围 left = mid + 1; } } //右边是有序的 if(nums[right] >= nums[mid]){ if(target > nums[mid] && target <= nums[right]){ left = mid+1; }else{ //右边有序但是都小于要查找的值 right = mid - 1; } } } return -1; }
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