33. Search in Rotated Sorted Array
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假设某个按照升序排序的数组在某个元素前旋转(例如:0 1 2 4 5 6 7 =>4 5 6 7 0 1 2)。给定一个目标值,判断该数组是否包含这个元素,如果包含返回对应的下标索引,否则返回-1.
解题思路:
可以发现,去中间值,总是存在一侧数据为有序数据,另一侧为无序数据。所以可以通过判断目标值是否在有序一侧的范围内,进行分开处理。
由于数组本来为有序数组,如果目标值在有序一侧的范围,无需考虑无序一侧;如果不在有序一侧,则考虑无序一侧的数据,再次进行二分查找。
实现:
int search(vector<int>& nums,int target){ int m = 0; int n = nums.size() - 1; int mid = 0; while(m <= n) { mid = (m + n) / 2; if(nums[mid] == target) return mid; if(nums[m] <= nums[mid]) { if(nums[m] <= target && nums[mid] > target) { n = mid - 1; } else { m = mid + 1; } } else { if(nums[mid] < target && nums[n] >= target) { m = mid + 1; } else { n = mid - 1; } } } return -1;}
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