Binary search LeetCode33 rotation sorted array

Search in Rotated Sorted Array

description

Suppose a sorted array, and then the rotation Pivot according to one, such as [1,2,3,4,5,6,7,8] -> [6,7,8,1,2,3,4 5].
Your task is to rotate find the array after a number in the subscript, subscript such as 6 to 1, and if this number does not exist, -1 is returned.
Requirements: time complexity of O (log (n))

solution

• Obviously we should expect to see logn divide and rule (Divide and Conquer). How to use the D & C There are three ways
  • 1. This problem easiest way is through divide and rule first find the pivot mentioned above, the "spin" Go back, then go to have this number.
  • 1. According to "spin" feature, in addition to the array will find somewhere step occurred, the other is increasing. Mid no matter where, there is always one side (left-> mid, mid-> right ) is monotonically increasing. So that we can always incrementally determine the relationship between the position of mid and target edges.
  • 1. For 2 further summarizes the improved method, nums[0]<=target, target<=nums[i], nums[i]<nums[0]the three conditions contains all the circumstances, and then XOR wrote a concise way: https://leetcode-cn.com/problems/search-in-rotated-sorted-array/ solution / ji-jian-solution- by-lukelee /

Boundary conditions

• Obviously, the above is not difficult to think of ideas, we thought at least the second step can be done, but the boundary conditions still need to think about.
• Boundary binary search algorithm, generally divided into two cases, one is left closed right-open interval, similar to the [left, right), one is left close right close range, similar to the [left, right]. It should be noted It is cyclic in vitro initialization conditions, the loop iteration step the body, must comply with the rules consistent interval.
• The best way to Programming Pearls given:

int search4(int array[], int n, int v)
{
int left, right, middle;

left = -1, right = n;

while (left + 1 != right)
{
    middle = left + （right － left) / 2;

    if (array[middle] < v)
    {
        left = middle;
    }
    else
    {
        right = middle;
    }
}

if (right >= n || array[right] != v)
{
    right = -1;
}

return right;
}

Reference: https://blog.csdn.net/weixin_43705457/article/details/87874779