！leetcode[33&81]:Search in Rotated Sorted Array[I & II]

最新推荐文章于 2020-06-05 15:27:31 发布

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本文深入解析了搜索旋转排序数组的问题，并针对允许重复元素的情况进行了拓展讨论，提供了相应的解决策略。

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.

int search(int* nums, int numsSize, int target) {
    int i;
    for(i=0;i<numsSize;i++)
    {
        if(nums[i]==target) return i;
    }
    return -1;
}

Search in Rotated Sorted Array II

Follow up for “Search in Rotated Sorted Array”:
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

bool search(int* nums, int numsSize, int target) {
    int i;
    for(i=0;i<numsSize;i++)
    {
        if(nums[i]==target) return true;
    }
    return false;

}