35. Search Insert Position 搜索插入位置

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.

Example 1 :

Input: [1,3,5,6], 5
Output: 2

Example 2 :

Input: [1,3,5,6], 2
Output: 1

Example 3 :

Input: [1,3,5,6], 7
Output: 4

Example 4 :

Input: [1,3,5,6], 0
Output: 0

Difficulty: Easy

解法1

二分法

class Solution {
public:
    int searchInsert(vector<int>& nums, int target) {
        if (nums.back() < target) return nums.size();
        int left = 0,right = nums.size()-1;
        while(left < right)
        {
            int mid = (left+right)/2;
            if (nums[mid]==target) return mid;
            else if (nums[mid]<target) left = mid+1;
            else right = mid;
        }
        return right;
    }
};

解法2

由于对于复杂度没有要求,遍历一次

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