zoukankan      html  css  js  c++  java
  • [leetcode]33. Search in Rotated Sorted Array旋转过有序数组里找目标值

    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

    题目

    一个有序数组,可能进行了循环移位,在里面进行查找。

    Solution1: Two Pointers(left&right)

    1. We can observe even rotating the array,  at least one part of such array is sorted.

    2. Find such sorted array part and do binary search

    3. how to find such sorted array?  

    (1) use two pointers: left, right 

    (2) if nums[left] < nums[mid],  we can say from left to mid, it is in ascending order, then left part is sorted

    (3) if nums[left] > nums[mid], we can say from left to mid, it is NOT in ascending order, then left part is NOT sorted

    code

     1 /*
     2 Time: O(logn). We use binary search
     3 Space:O(1) . We only used constant extra space. 
     4 */
     5 
     6 class Solution {
     7     public int search(int[] nums, int target) {
     8         // corner case
     9         if (nums.length == 0) return -1;
    10         // initialize
    11         int left = 0;
    12         int right = nums.length - 1;
    13 
    14         while (left <= right) {
    15             int mid = left + (right - left) / 2;
    16             if (nums[mid] == target) return mid;
    17             // left part is not in ascending order
    18             else if (nums[mid] < nums[left]) { 
    19                 if (target > nums[mid] && target <= nums[right]) {
    20                     left = mid + 1;
    21                 } else {
    22                     right = mid - 1;
    23                 }
    24             } 
    25             // nums[mid] >= nums[left] left side is in ascending order
    26             else { 
    27                 if (target < nums[mid] && target >= nums[left]) {
    28                     right = mid - 1;
    29                 } else {
    30                     left = mid + 1;
    31                 }
    32             }
    33         }
    34         return -1;
    35     }
    36 }
  • 相关阅读:
    Apple MDM
    苹果核
    iOS自动化测试的那些干货
    Wifi 定位原理及 iOS Wifi 列表获取
    详解Shell脚本实现iOS自动化编译打包提交
    PushKit 占坑
    【译】使用 CocoaPods 模块化iOS应用
    NSMutableArray 根据key排序
    iOS 通过tag查找控件
    自己使用 2.常量变量,数据类型,数据的输入输出。
  • 原文地址:https://www.cnblogs.com/liuliu5151/p/9820406.html
Copyright © 2011-2022 走看看