Binary Search:33. Search in Rotated Sorted Array
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这道题让在旋转数组中搜索一个给定值,若存在返回坐标,若不存在返回-1。
采用二分搜素,找到nums[mid]位置,如果nums[mid]>nums[left],说明left到mid是有序的,那么看target是否在left到mid这一段,在的话right = mid - 1,不在的话left = mid + 1。如果mid到right部分是有序的,那么看target是否在mid到right这一段,在的话left = mid + 1,不在的话right = mid - 1;
class Solution {public: int search(vector<int>& nums, int target) { int left = 0, right = nums.size()-1; while(left <= right) { int mid = left + (right - left) / 2; if(nums[mid] == target) return mid; if(nums[mid] > nums[left]) { if(target >= nums[left] && target <= nums[mid]) right = mid - 1; else left = mid + 1; } else if(nums[mid] < nums[left]) { if(target >= nums[mid] && target <= nums[right]) left = mid + 1; else right = mid - 1; } else left++; } return -1; }};
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