思路:有序查找首先想到二分;这里的话,再加一个判断:如果最后得到的下标对应的值小于target的话,说明target应该插入到这个下标的右边;否则,应该插入到该下标位置;
class Solution {
public:
int searchInsert(vector<int> &nums, int target)
{
//binary_Search
int left = 0, right = nums.size() - 1;
while (left < right)
{
int mid = (left + right) / 2;
if (nums[mid] == target)
{
return mid;
}
else if (target > nums[mid])
{
left = mid + 1;
}
else
{
right = mid - 1;
}
}
if (nums[left] < target) //if nums[left] < target, we should insert it to the right, so return left+1;
{
return left + 1;
}
return left;
}
};