[C语言][LeetCode][33]Search in Rotated Sorted Array

题目

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.

标签

Array、Binary Search

难度

难

分析

题目意思是一个旋转后的排序数组，让你查找一个value值，如果value存在则返回index，否则返回-1。解题思路是利用二分法去解决，分开两段，如有有一段不是排好序的，那说明另一段肯定是排好序的。主要有下面三种情况：
1）4567012：left＜mid 、mid＞right
2）0124567：left＜mid、mid＜right
3）7012456：left＞mid、mid＜right

由上可知，当left＜mid的时候，左半部分肯定是排好序的，右半部分要么是排好序的，要么是没有排好序的。

C代码实现

int search(int* nums, int numsSize, int target) 
{
    int l = 0;
    int r = numsSize-1;

    while(l<=r)
    {
        int mid = l + (r-l)/2;
        if(target == nums[mid])
            return mid;

        if(nums[l]<=nums[mid])//it is sorted in [l,mid], "=" contains an edge case:3, 1
        {
            if(nums[mid] > target && nums[l] <= target)  
                    r = mid-1;  
                else   
                    l = mid+1;
        }
        else//[mid,r] must be sorted, if [l,mid] not sorted 
        {  
            if(nums[mid] < target && nums[r] >= target)  
                l = mid+1;  
            else   
                r = mid-1;  
        }

    }
    return -1;
}