题解 | #在旋转过的有序数组中寻找目标值#
在旋转过的有序数组中寻找目标值
http://www.nowcoder.com/practice/87c0e7abcbda41e7963660fa7d020995
class Solution { public: /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param nums int整型vector * @param target int整型 * @return int整型 */ int search(vector<int>& nums, int target) { int left = 0; int right = nums.size() - 1; while(left<=right){ int mid = left + (right - left)/2; if(nums[mid] == target){ return mid; } if(nums[mid] < nums[right]){ if(nums[mid] < target && nums[right] >= target){ left = mid + 1; } else { right = mid - 1; } } else{ if(nums[left] <= target && nums[mid] > target){ right = mid - 1; } else { left = mid + 1; } } } return -1; } // write code here };左闭右闭 , 中间值与右侧值比较,小于右侧值即右侧有序,在有序部分二分查找https://www.cnblogs.com/grandyang/p/4325648.html