35. Search Insert Position

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题目

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

思路

这个应该算easy吧……不是很懂难度划分

找到大于等于target的数字返回下标就好。

代码

暴力……

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//C++
class Solution {
public:
    int searchInsert(vector<int>& nums, int target) {
        for(int i=0;i<nums.size();i++){
            if(nums[i]>=target){
                return i;
            }
            else if(i==nums.size()-1){
                return i+1;
            }
        }
        return 0;
    }
};

二分法：

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//C++
class Solution {
public:
    int searchInsert(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;
            }
            else if(nums[mid]<target){
                left = mid + 1;
                
            }
            else {
                right = mid - 1;
                
            }
        }
        return right + 1;
    }
};