Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume no duplicates in the array.
Here are few examples.[1,3,5,6]
, 5 → 2[1,3,5,6]
, 2 → 1[1,3,5,6]
, 7 → 4[1,3,5,6]
, 0 → 0
class Solution { public: //二分查找,找不到的时候如果大于target,就返回left,如果小于target,就加一位 int searchInsert(vector<int>& nums, int target) { int left = 0; int right = nums.size() - 1; int mid; while (left < right) { mid = left + (right - left) / 2; if (nums[mid] == target) { return mid; } else if (nums[mid] < target) { left = mid + 1; } else { right = mid - 1; } } if (nums[left] >= target) { return left; } else { return left + 1; } } };