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  • Leetcode: Search in Rotated Sorted Array

    Suppose a sorted array 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.

    根据binary search,每次我们都可以切掉一半的数据,所以算法的时间复杂度是O(logn),空间复杂度是O(1)。

     1 class Solution {
     2     public int search(int[] nums, int target) {
     3         if (nums == null || nums.length == 0) return -1;
     4         int l = 0, r = nums.length - 1;
     5         while (l <= r) {
     6             int m = l + (r - l) / 2;
     7             if (nums[m] == target) return m;
     8             if (nums[m] >= nums[l]) {
     9                 // [l, m] is in order
    10                 if (nums[l] <= target && target < nums[m]) {
    11                     r = m - 1;
    12                 }
    13                 else l = m + 1;
    14             }
    15             else {
    16                 // [m, r] is in order
    17                 if (nums[m] < target && target <= nums[r]) {
    18                     l = m + 1;
    19                 }
    20                 else r = m - 1;
    21             }
    22         }
    23         return -1;
    24     }
    25 }
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  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/3978881.html
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