• 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

    this is my first code, it can't satisfy the last condition:

    class Solution {
    public:
        int search(vector<int>& nums, int target) {
            int len = nums.size();
            if (len == 0) return -1;
            bool flag = false;
            if (target < nums[0]) {
                for (int i = len-1; i > 0; --i) {
                    if (nums[i] == target) {
                        flag = true;
                        return i;
                    } 
                }
            } else {
                for (int i = 0; i < len; ++i) {
                    if (nums[i] == target) {
                        flag = true;
                        return i;
                    }
                        
                }
            }
            if (!flag) {
                return -1;
            }
        }
    };

    this is the right way to use binary search:

    class Solution {
    public:
        int search(vector<int>& nums, int target) {
            int len = nums.size();
            if (len == 0) return -1;
            int l = 0; 
            int r = len - 1;
            while (l <= r) {
                int m = (l + r) / 2;
                if (nums[m] == target) return m;
                if (nums[m] > nums[r]) {
                    if (target < nums[m] && target >= nums[l]) {
                        r = m - 1;
                    } else {
                        l = m + 1;
                    }
                } else if (nums[m] < nums[l]) {
                    if (target <= nums[r] && target > nums[m]) {
                        l = m + 1;
                    } else {
                        r = m -1;
                    }
                } else {
                    if (target < nums[m]) {
                        r = m -1;
                    } else {
                        l = m + 1;
                    }
                }
            }
            return -1;
        }
    };
    

    Runtime: 8 ms, faster than 25.99% of C++ online submissions for Search in Rotated Sorted Array.

    but this way cost more time before the previous way.

    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/9775497.html
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