Search Insert Position


  • 0
    F

    a convenient way to solve it just in this thinking:find the first value which is larger than the target,and the index of the value is the answer.

    int searchInsert(int* nums, int numsSize, int target) {
        if(numsSize == 0) return 0;
        int j=-1;
        for(int i=0;i<numsSize;i++)
        {
            if(nums[i]>=target)
            {
                j=i;
                break;
            }
        }
        if(j==-1) j=numsSize;
        return j;
    }
    

    The time cost is O(n),The space cost is O(1)


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