33. Search in Rotated Sorted Array(js)

33. Search in Rotated Sorted Array

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
题意:局部排好序的数组,匹配target值,返回索引,未匹配到则返回-1
代码如下:
/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
//二分查找
var search = function(nums, target) {
       let left=0,right=nums.length-1;
        while(left<=right){
            let mid=left+parseInt((right-left)/2);
            if(nums[mid]==target) return mid;
            else if(nums[mid]<nums[right]){
                if(nums[mid]<target && nums[right]>=target) left=mid+1;
                else right=mid-1;
            }else{
                if(nums[mid]>target && nums[left]<=target) right=mid-1;
                else left=mid+1;
            }
        }
        return -1;
};
原文地址:https://www.cnblogs.com/xingguozhiming/p/10409313.html